NPTEL Introduction to programming in C Programming Assignments Jan-2024 Swayam

 

NPTEL » Introduction to programming in C

Jan 2024 NPTEL course

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Week 1: Assignment 1 - Question 1

Due on 2024-02-08, 23:59 IST

Problem Statement

You have a certain number of 100 rupee notes, 10 rupee notes and 1 rupee notes with you.

There is an item you want to buy whose price is given to you.

Write a program to find if the item is affordable, that is the price of the item is less than or equal to the current money you have.

Input

Four non negative integers. 

The first input is an integer representing the number of 100 rupee notes.

The second input is an integer representing the number of 10 rupee notes.

The third input is an integer representing the number of 1 rupee notes.

The fourth input is an integer representing the price of the item.

Output

You have to output 1 if the item is affordable.

You have to output 0 if the item is not affordable.


Sample Inputs and Outputs

Sample input 1
--------------------

9 10 1 1001

Sample Output 1
----------------------

1

Explanation
----------------

The total amount of money is 900 + 100 + 1 = 1001, which is equal to the price of the item.
So, the item is affordable.

Sample input 2
--------------------

9 10 0 1001

Sample Output 2
----------------------

0

Explanation
----------------

The total amount of money is 900 + 100 + 0 = 1000, which is less than the price of the item.
So, the item is not affordable.

Your last recorded submission was on 2024-02-04, 10:41 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

int main()

3

{

4

    int hundreds,tens,ones;

5

    int price;

6

    int money;

7

8

    scanf("%d",&hundreds);

9

    scanf("%d",&tens);

10

    scanf("%d",&ones);

11

    scanf("%d",&price);

12

13

    money = hundreds*100 + tens*10 + ones;

14

15

    if( price <= money){

16

        printf("1");

17

    }

18

    else{

19

        printf("0");

20

    }

21

22

    return 0;

23

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	9 10 1 1001	1	1	Passed
Test Case 2	9 10 0 1001	0	0	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed
Test Case 4	Passed
Test Case 5	Passed

Week 1: Assignment 1 - Question 2

Due on 2024-02-08, 23:59 IST

Problem Statement

Given three distinct integers a b and c, write a C program to find the second largest number among them.

Input

Three distinct integers a b c.

The first input is the integer a.

The second input is the integer b.

The third input is is the integer c.

Output

The second largest among a, b and c

Sample Inputs and Outputs

Sample input 1
--------------------

9 10 1 

Sample Output 1
----------------------

9

Explanation
----------------

1 < 9 < 10

Sample input 2
--------------------

-4 3 9

Sample Output 2
----------------------

3

Explanation
----------------

-4 < 3 < 9

Your last recorded submission was on 2024-02-04, 10:43 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

int main(){

3

  int a,b,c;

4

  scanf("%d %d %d", &a, &b, &c);

5

  if( a > b )

6

    if(a > c)

7

        if( c > b )

8

            printf("%d", c);

9

        else

10

            printf("%d", b);

11

    else 

12

        printf("%d", a);

13

  else

14

    if(b > c)

15

        if ( c > a )

16

            printf("%d", c);

17

        else

18

            printf("%d", a);

19

    else

20

        printf("%d", b);

21

  return 0;

22

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	9 10 1	9	9	Passed
Test Case 2	-1 3 9	3	3	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed
Test Case 4	Passed
Test Case 5	Passed

Week 1: Assignment 1 - Question 3

Due on 2024-02-08, 23:59 IST

Problem Statement

Given the coefficients of a pair of linear equations,

a1x+b1y=c1a2x+b2y=c2$$\begin{gathered} {�}_1�+{�}_1�={�}_1 \\ {�}_2�+{�}_2�={�}_2 \end{gathered}$$

Find the solutions to x$�$ and y$�$

Input

Input consists two lines.
The first line contains coefficients of first equation,  a1b1c1${�}_1{�}_1{�}_1$  in that order

The second line contains coefficients of second equation,  a2b2c2${�}_2{�}_2{�}_2$ in that order

Output

The solutions to x and y.

Note : You can assume that both x and y will always be integers.
You can also assume that the solution is unique.

Sample input 
--------------------

2 3 7
4 -2 6

Sample Output 
----------------------

2 1 


Explanation
----------------

The set of equations are :

2x+3y=74x−2y=6$$\begin{gathered} 2�+3�=7 \\ 4�-2�=6 \end{gathered}$$

The solution is x=2;y=1.$�=2;�=1.$


Hint

Recall the gaussian elimination method learned in high school.

For example consider the set of equations,

2x+3y=74x−2y=6$$\begin{gathered} 2�+3�=7 \\ 4�-2�=6 \end{gathered}$$

To solve for x, multiply the first equation by b2=−2${�}_2=-2$ and the second equation by b1=3${�}_1=3$, we get 

−4x−6y=−1412x−6y=18$$\begin{gathered} -4�-6�=-14 \\ 12�-6�=18 \end{gathered}$$

Subtracting equation 1 and 2, we get −16x=−32$-16�=-32$ and therefore x=−32−16=2.$�=\frac{-32}{-16}=2.$.

One can similarly solve for y.

Formulate the steps involved during gaussian elimination in terms of a1 b1 c1 and a2 b2 c2 and convert this into a program.

Your last recorded submission was on 2024-02-04, 10:44 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

int main(){

3

  int a1,b1,c1,a2,b2,c2;

4

  scanf("%d %d %d", &a1, &b1, &c1);

5

  scanf("%d %d %d", &a2, &b2, &c2);

6

7

  int x = (b2*c1-c2*b1)/(b2*a1-b1*a2);

8

  int y = (a2*c1-c2*a1)/(a2*b1-a1*b2);

9

10

  printf("%d %d", x, y);

11

12

  return 0;

13

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 1 / 1 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	2 3 7 4 -2 6	2 1	2 1	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 2: Assignment 2 - Question 1

Due on 2024-02-08, 23:59 IST

Parity Checker

You are given a sequence of bits (1's and 0's).
The sequence is said to have even parity if and only if the number of 1's in the sequence if even.

Write a C program to that outputs 1 if the sequence has even parity and 0 otherwise.

Input

A sequence of bits (0's and 1's) ending with a -1.
(Note : -1 is not a part of input. It only signifies that input has ended)

Output

1 if the number of ones in the sequence is even.
0 if the number of ones in the sequence is odd.

Sample Inputs and Outputs

Sample input 1
--------------------

0 1 1 -1

Sample Output 1
----------------------

1  

Explanation
----------------
The input sequence is  011$011$ . It has two ones. Since 2 is even, the sequence has even parity.


Sample input 2
--------------------

0 1 0 1 0 1 -1

Sample Output 1
----------------------

0  

Explanation
----------------
The input sequence is 010101$010101$ . The number of 1's is three. Since 3 is odd,  the sequence does not have even parity.

Your last recorded submission was on 2024-02-04, 10:50 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

int main()

3

{

4

  int b, count = 0;

5

  scanf("%d", &b);

6

  while ( b!= -1)

7

  {

8

    if( b == 1)

9

      count++;

10

    scanf("%d", &b);

11

  }

12

  printf("%d", count%2==0);

13

  return 0;

14

}    

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	0 1 1 -1	1	1	Passed
Test Case 2	0 1 0 1 0 1 -1	0	0	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

There is no Question 2 provided in the assignments from the NPTEL course team.

Week 2: Assignment 2 - Question 3

Due on 2024-02-08, 23:59 IST

Count the Number of 0's Between the First and Last 1

You are given a binary sequence.

Write a C program to count the number of 0's between the first and last 1 in the sequence.

Input

A sequence of bits (0's and 1's) ending with a -1.
(Note : -1 is not a part of input. It only signifies that input has ended)

Output

The number of 0's Between the First and Last 1 in the sequence.

Note : Make no assumptions about the data in the sequence.
For instance if there is no starting and ending 1 ( say the sequence is all 0), you have to output 0. 

Sample Inputs and Outputs

Sample input 1
--------------------

0 1 0 0 1 1 0 1 0 0 -1

Sample Output 1
----------------------

3  

Explanation
----------------

The sequence between first and last 1 is 1 0 0 1 1 0 1 and the sequence has 3 zeroes in between.

Sample input 2
--------------------

0 0 0 1 1 1 1 1 1 0 0 -1

Sample Output 2
----------------------

0  

Explanation
----------------

The sequence between first and last 1 is 1 1 1 1 1 1 and the sequence has 0 zeroes in between.

Sample input 3
--------------------

0 0 1 0 0 0 -1

Sample Output 1
----------------------

0 

Explanation
----------------

The sequence has a single 1, therefore the output (by problem description) is 0.

Your last recorded submission was on 2024-02-04, 10:52 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

int main()

3

{

4

  int bit, count = 0, flag = 0, sum = 0;

5

  scanf("%d", &bit);

6

  while ( bit!= -1)

7

  {

8

    if( bit == 1)

9

    {flag = 1;

10

      count+=sum;

11

      sum = 0;

12

     scanf("%d", &bit);}

13

    if( flag )

14

      if( bit == 0)

15

      sum++;

16

      scanf("%d", &bit);

17

  }

18

  printf("%d", count);

19

  return 0;

20

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 4 / 4 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	0 1 0 0 1 1 0 1 0 0 -1	3	3	Passed
Test Case 2	0 0 0 1 1 1 1 1 1 0 0 -1	0	0	Passed
Test Case 3	0 0 0 1 0 1 1 0 1 0 0 -1	2	2	Passed
Test Case 4	0 0 1 0 0 0 -1	0	0	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 3: Assignment 3 - Question 1

Due on 2024-02-15, 23:59 IST

Find moving average

In this question, you have to output the "two moving average" of a sequence of non-negative numbers.

The two moving average is the sequence of averages of the last 2 entries.
For the first number, no average is output.

For example, if the sequence of numbers is a1,a2,a3,a4,a5

The 2-moving average is  (a1+a2)2,(a2+a3)2,(a3+a4)2,(a4+a5)2

Input

-------

The input is a sequence of non-negative numbers, terminated by a -1.
There will be at least 3 numbers in the sequence.

Note: The -1 is not part of the sequence. It is just to indicate that the input has ended.

Output

----------

You have to output the moving average of the sequence. The output should be printed correct to one digit after the decimal.

Hint : Use the format specifier "%.1f" inside printf.

Sample Input 1

---------------------

1 3 2 4 -1

Sample Output 1

-------------------------

2.0 2.5 3.0

Explanation
-------------------------
(1+3)/2 = 2
(3+2)/2 = 2.5
(2+4)/2 = 3

Your last recorded submission was on 2024-02-06, 15:41 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

void main()

3

{

4

int count=1;

5

float x0,x1;

6

7

    scanf("%f",&x0);

8

    while(x0!=-1)

9

    {

10

          scanf("%f",&x1);

11

          if(x1!=-1 && count)

12

            printf("%.1f",(x0+x1)/2);

13

          if(x1!=-1 && !count)

14

            printf(" %.1f",(x0+x1)/2);

15

      count=0;

16

      x0=x1;

17

    }

18

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	1 3 2 4 -1	2.0 2.5 3.0	2.0 2.5 3.0	Passed
Test Case 2	1 3 2 4 -1	2.0 2.5 3.0	2.0 2.5 3.0	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 3: Assignment 3 - Question 2

Due on 2024-02-15, 23:59 IST

Prime Checking

Complete the function int is_prime(int n) to check if a positive number n is prime or not.

The function returns 1 if n is prime, and 0 otherwise.

The function will be used in a program (code given) that prints the prime numbers in a given sequence.

Input

-------

The first line of input is a positive integer N.
The next line contains N positive integers ki for i=1 to N.

Output
---------
The elements in the input list which are primes, in the original order.

Sample input
------------------
12
2 3 4 5 6 7 8 9 10 11 12 13

Sample output
--------------------
2 3 5 7 11 13

Note: Ignore the "Passed after ignoring Presentation Error" comment.

Your last recorded submission was on 2024-02-06, 15:42 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

​

3

int is_prime(int n){

4

  int i,r;

5

  for(i=n/2;i>1;i--)

6

  {

7

    r=n%i;

8

    if(r==0)

9

        return 0;

10

  }

11

  return 1;

12

}

13

​

14

int main() {

15

​

16

  int n,num;

17

  scanf("%d", &n);

18

​

19

  for(int i=0;i<n;i++){

20

      scanf("%d",&num);

21

      if(is_prime(num)){

22

          printf("%d ",num);

23

      }

24

  }

25

​

26

  return 0;

27

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 1 / 1 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	12 2 3 4 5 6 7 8 9 10 11 12 13	2 3 5 7 11 13	2 3 5 7 11 13	Passed after ignoring Presentation Error

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 3: Assignment 3 - Question 3

Due on 2024-02-15, 23:59 IST

Find the kth Odd integer in sequence

Write a C function to find the kth occurrence of an odd integer in a sequence of non-

negative integers.

Input
--------

You are given the input in two lines:

• The first line contains a positive integer k.
  
  
• In the second line, you will be given a sequence of numbers terminated with a -1.

You have to find the kth occurrence of an odd integer in the sequence.

Note:  The -1 is not part of the sequence.

Output
----------

If there are k odd numbers in the sequence, then output the kth occurrence of an odd number in the sequence, if present. If there are less than k odd numbers in the sequence, output -1.

Sample Input
------------------

2

1 4 8 3 6 5 2 3 4 1 -1

Sample Output
--------------------

3

Explanation
-------------------
The odd integers in the sequence in order are :
1 3 5 3 1

The second odd integer in the list is therefore 3. 

Your last recorded submission was on 2024-02-06, 15:44 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

int find_odd(int k);

3

void main()

4

{

5

    int k;

6

    scanf("%d",&k);

7

    printf("%d",find_odd(k));

8

9

}

10

int find_odd(int k){

11

  int x,cnt=0, flag=1;

12

  scanf("%d",&x);

13

    while(x!=-1)

14

    {

15

          if(x%2!=0)

16

                ++cnt;

17

          if(cnt==k)

18

          { return x;

19

          }

20

      scanf("%d",&x);

21

    }

22

  if (flag)

23

    return -1;

24

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 1 / 1 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	2 1 4 8 3 6 5 2 3 4 1 -1	3	3	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 4: Assignment 4 - Question 1

Due on 2024-02-22, 23:59 IST

Description

Given an array of positive integers, write a C Program to output the sum of the elements that are above the mean.

Given an array [a1,a2…an]$[{�}_1,{�}_2\dots {�}_{�}]$, the mean of the array is defined as μ=∑ni=1ain$�=\frac{\sum _{�=1}^{�}{�}_{�}}{�}$.

Input

The first line contains the size of the array.
The second line contains the contents of the array. 

Output 

Output the sum of elements in the array which are greater than or equal to the mean.
  

Note : The size of the array is always smaller than 20.

Sample input 1
--------------------

5
1 2 3 4 5

Sample Output 1
----------------------

12  

Explanation
----------------
The mean of the array is (1+2+3+4+5)/5 = 3.
The elements 3,4,5 are greater than or equal to 3, and 3+4+5 = 12.

Sample input 2
--------------------

8
1 8 2 1 6 5 9 4

Sample Output 2
----------------------

28  

Explanation
----------------
The mean is 4.5 and the numbers greater than or equal to this value in the array are 8,6,5,9 
8+6+5+9 = 28.

Your last recorded submission was on 2024-02-10, 10:35 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

​

3

#define SIZE 25

4

int main() {

5

  int arr[SIZE];

6

  int i;

7

  int j;

8

  int sum = 0;

9

  float mean;

10

  int n;

11

  scanf("%d", &n);

12

13

  for (i = 0; i < n; i++) {

14

    scanf("%d", &arr[i]);

15

    sum+=arr[i];

16

  }

17

18

  mean = (float)sum/n;

19

  sum=0;

20

  for (i = 0; i < n; i++) 

21

      if (arr[i] >= mean)

22

          sum+=arr[i];

23

  printf("%d",sum);

24

  return 0;

25

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	5 1 2 3 4 5	12	12	Passed
Test Case 2	8 1 8 2 1 6 5 9 4	28	28	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 4: Assignment 4 - Question 2

Due on 2024-02-22, 23:59 IST

Description

Given two arrays of positive integers, write a C Program to output the smallest number in the first array that is also present in the second one.

Input

The first line contains the size of the first array.
The second line contains the contents of first array. 
The third line contains the size of the second array.

The fourth line contains the contents of second array. 

Output 

Output the smallest number occurring in the first array that also occurs in the second.
In case there is no such number, output -1.

Note : The sizes of the arrays are smaller than 20.

Sample input 1
--------------------

3
2 4 3
4
1 3 6 4

Sample Output 1
----------------------

3  

Explanation
----------------
The numbers in the first array that also appear in the second array are 3 , 4 . 
The smallest among them is 3.

Sample input 2
--------------------

4
1 8 2 1
4
6 5 9 3

Sample Output 2
----------------------

-1  

Explanation
----------------
No numbers from the first array appear in the second array. 
The output is -1.

Your last recorded submission was on 2024-02-10, 10:57 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

​

3

#define MAX 20

4

int read_array(int arr[]) {

5

  int i,n;

6

  scanf("%d",&n);

7

  for (i = 0; i < n; i++)

8

    scanf("%d", & arr[i]);

9

  return n;

10

}

11

int present(int arr[], int n, int elt) {

12

  int i;

13

  for (i = 0; i < n; i++) {

14

    if (arr[i] == elt) {

15

      return 1;

16

    }

17

  }

18

  return 0;

19

}

20

int main() {

21

  int n1,n2, flag = 1;

22

  int arr1[MAX];

23

  int arr2[MAX];

24

25

  n1 = read_array(arr1);

26

  n2 = read_array(arr2);

27

28

  int i, smallest = 36727;

29

30

  for (i = 0; i < n1; i++) {

31

    int val = arr1[i];

32

    if (present(arr2, n2, val)) {

33

      if (smallest > val) {

34

        smallest = val;

35

        flag = 0;

36

      }

37

    }

38

  }

39

40

  if (flag)

41

    printf("%d", -1);

42

  else

43

    printf("%d", smallest);

44

​

45

  return 0;

46

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	3 2 4 3 4 1 3 6 4	3	3	Passed
Test Case 2	4 1 8 2 1 4 6 5 9 3	-1	-1	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed
Test Case 4	Passed

Week 4: Assignment 4 - Question 3

Due on 2024-02-22, 23:59 IST

Anagram Checking

Given two strings(character arrays), write a C Program to check if one of them is an anagram of the other.

An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, using all the original letters exactly once.

Examples are LISTEN and SILENT , KNEE and KEEN.

Input

The first line contains the size of the character array.
The second line contains the contents of first array. 
The third line contains the size of the second array.

Note: The maximum size of the character array can be assumed to be 20.
The contents of both arrays are only upper case alphabets.

Output 

1 If the contents of the arrays are anagrams
0 If the contents of the arrays are not anagrams

Note : The sizes of the both the arrays can be assumed to be the same.

Sample input 1
--------------------

6
SILENT
LISTEN

Sample Output 1
----------------------

1  

Explanation
----------------
SILENT and LISTEN are anagrams.

Sample input 2
--------------------

4
FARE
SAFE

----------------------

0  

Explanation
----------------
FARE and SAFE are not anagrams.
FARE has an 'R' and does not contain 'S' from SAFE.

Your last recorded submission was on 2024-02-10, 12:25 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

​

3

#define MAX 20

4

int read_array(char arr[]) {

5

  int i,n;

6

  scanf("%d",&n);

7

  for (i = 0; i < n; i++)

8

    scanf(" %c", &arr[i]);

9

  return n;

10

}

11

​

12

void lexicographical_sort(char arr[], int n) {

13

  int i,j;

14

  char temp;

15

  for (i = 0; i < n-1; i++) {

16

    for (j = i+1; j < n; j++)

17

      if (arr[i] > arr[j]) {

18

        temp = arr[j];

19

        arr[j] = arr[i];

20

        arr[i] = temp;

21

      }

22

  }    

23

}

24

​

25

int main() {

26

  int n1,n2,i;

27

  char arr1[MAX];

28

  char arr2[MAX];

29

30

  n1 = read_array(arr1);

31

  n2 = read_array(arr2);

32

​

33

  lexicographical_sort(arr1,n1);

34

  lexicographical_sort(arr2,n2);

35

​

36

  for (i = 0; i < n1; i++) 

37

    if ( arr1[i] == arr2[i] )

38

      continue;

39

    else

40

    {

41

      printf("0");

42

      return 0;

43

    }

44

  printf("1");

45

  return 0;

46

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 3 / 3 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	6 LISTEN SILENT	1	1	Passed
Test Case 2	4 FARE SAFE	0	0	Passed
Test Case 3	4 KEEP KELP	0	0	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 5: Assignment 5 - Question 1

Due on 2024-02-29, 23:59 IST

Convergence depth of Collatz function

The Collatz function is defined for a positive integer n as follows.
f(n)={3n+1,n/2,if n oddif n is even

We consider the repeated application of the Collatz function starting with a given integer n, as

follows: 

f(n), f(f(n)), f(f(f(n))),…

It is conjectured that no matter which positive integer n you start from, this sequence eventually will have 1 in it. It has been verified to hold for numbers up to 5 × 260 [Wikipedia: Collatz Conjecture].

e.g. If n=7, the sequence is

1.     f(7) = 22
2.     f(f(7)) = f(22) = 11
3.     f(f(f(7))) = f(11) = 34
4.     f(34) = 17
5.     f(17) = 52
6.     f(52) = 26
7.     f(26) = 13
8.     f(13) = 40
9.     f(40) = 20
10.     f(20) = 10
11.     f(10) = 5
12.     f(5) = 16
13.     f(16) = 8
14.     f(8) = 4
15.     f(4) = 2
16.     f(2) = 1

Thus if you start from n=7, you need to apply f 16 times in order to first get 1.

In this question, you will be given a positive number <= 32,000. You have to output how many

times f has to be applied repeatedly in order to first reach 1.


Sample Input
------------------

7


Sample Output
--------------------

16

Your last recorded submission was on 2024-02-19, 17:46 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

​

3

int f(int n){

4

  static int count = 0;

5

  if( n == 1 )

6

    return count;

7

  else {

8

    if(n%2)

9

      f(3*n + 1);

10

    else

11

      f(n/2);

12

  count++;

13

  }

14

}

15

​

16

int main(){

17

  int n;

18

  scanf("%d",&n);

19

  printf("%d",f(n));

20

  return 0;

21

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	7	16	16	Passed
Test Case 2	13	9	9	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 5: Assignment 5 - Question 2

Due on 2024-02-29, 23:59 IST

BlockSum of an Array

Given an integer array M$�$ having size n which is power of 2, 
Write a recursive code to find the BlockSum of the array M.

The following is the recursive  definition of BlockSum:

If size of M is 2, say M=[a,b]$�=[�,�]$, where a and b are integers, then BlockSum(M)=a−b$��������(�)=�-�$.

Otherwise (when n > 2), partition M into two subarrays of equal size:

                          M=[AB]$�=[��]$

The BlockSum of M is defined recursively as : 

                           BlockSum(M)=BlockSum(A)−BlockSum(B)$��������(�)=��������(�)-��������(�)$

Here A and B are arrays of Size n/2 each.
A is the first n/2 elements of M ( in the same order) and B is the last n/2 elements of M ( in the same order). 

Note : You can assume that size of input array is a power of 2, and the size is less than 1024. 

Input

The first line contains the array size n
The next n lines contains the elements of the array M.

Output

BlockSum(M)$��������(�)$

Sample Inputs and Outputs

Sample Input 1
--------------------

2
3 2

Sample Output 1
----------------------

1

Sample Input 2
--------------------

2
7 1 

Sample Output 2
----------------------

6

Sample Input 3
--------------------

4
7 1 3 2

Sample Output 3
----------------------

5

Explanation
--------------------

BlockSum( [ 3, 2 ] ) = 3 - 2  = 1

BlockSum( [ 7, 1 ] ) = 7 - 1  = 6

BlockSum( [ 7 , 1 , 3,  2 ] ) = BlockSum( [ 7, 1 ] )  - BlockSum( [ 3, 2 ] )  = 6 - 1 = 5

Your last recorded submission was on 2024-02-19, 18:17 IST

Select the Language for this assignment.      --                     C               

1

#include<stdio.h>

2

​

3

int BlockSum(int a[], int size){

4

  if( size == 2 )

5

    return a[0]-a[1];

6

  else

7

    return BlockSum(a, size/2) - BlockSum((a+(size/2)), size/2);

8

}

9

​

10

int main(){

11

  int size;

12

  scanf("%d",&size);

13

  int a[size];

14

  for(int i = 0; i<size; i++)

15

    scanf("%d",&a[i]);

16

  printf("%d",BlockSum(a, size));

17

  return 0;

18

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	2 3 2	1	1	Passed
Test Case 2	4 7 1 3 2	5	5	Passed

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed
Test Case 4	Passed

Week 6: Assignment 6 - Question 1

Due on 2024-03-07, 23:59 IST

ℓ-window smoothing.

Given an n×n integer Matrix A and an positive number ℓ such that 2ℓ+1≤n, write a C program to print the ℓ window smoothing of A. 

To get the  ℓ-window smoothing of A , we replace A[i][j] with the sum of the values of the imaginary submatrix S of A with centre at A[i][j], and having size 2ℓ+1×2ℓ+1

More precisely, the smoothed matrix B[i,j]=∑u=ilih∑v=jljhA[u][v]  
where il=max(i−ℓ,0),ih=min(i+ℓ,n−1),jl=max(j−ℓ,0),jh=min(j+ℓ,n−1).

Input
The first line contains the dimension of the matrix n. Assume n < 100.
The second line contains the smoothing parameter ℓ. 
The next n lines contains the contents of the matrix A, each row per line.

Output
The smoothed matrix of A

Note: Ignore the Passed after ignoring Presentation Error comment.

Example 

Input

411471258236934714

Output

12272718214545302745452720303017

Explanation

A[0][0] = 1 + 2 + 4 + 5 = 12 
A[1][1] = 1 + 2 +3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Your last recorded submission was on 2024-02-25, 12:17 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

​

3

int max(int a, int b){

4

    if(a>b)

5

        return a;

6

    return b;

7

}

8

​

9

int min(int a, int b){

10

    if(a<b)

11

        return a;

12

    return b;

13

}

14

​

15

int main() {

16

    int A[100][100];

17

    int B[100][100];

18

19

    int n,l,sum;

20

21

    scanf("%d",&n);

22

    scanf("%d",&l);

23

24

    for(int i = 0; i< n;i++){

25

        for(int j = 0; j< n;j++)

26

            scanf("%d",&A[i][j]);

27

    }

28

29

    for(int i = 0; i< n;i++){

30

        for(int j = 0; j< n;j++){

31

32

            int ih,il,jh,jl;

33

            sum =0;

34

35

            /*

36

               Code for calculating B[i][j]

37

           */

38

          for(il = max(i-l,0) ; il <= min(i+l,n-1) ; il++ )

39

            for(jl = max(j-l,0) ; jl <= min(j+l,n-1) ; jl++)

40

              sum += A[il][jl];

41

            B[i][j] = sum;

42

        }

43

    }

44

45

    for(int i = 0; i< n;i++){

46

        for(int j = 0; j< n;j++){

47

            printf("%d ",B[i][j]);

48

        }

49

        printf("\n");

50

    }

51

    return 0;

52

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	4 1 1 2 3 4 4 5 6 7 7 8 9 1 1 2 3 4	12 21 27 20 \n 27 45 45 30 \n 27 45 45 30 \n 18 30 27 17	12 21 27 20 \n 27 45 45 30 \n 27 45 45 30 \n 18 30 27 17 \n	Passed after ignoring Presentation Error
Test Case 2	6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1	4 6 6 6 6 4 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 4 6 6 6 6 4	4 6 6 6 6 4 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 6 9 9 9 9 6 \n 4 6 6 6 6 4 \n	Passed after ignoring Presentation Error

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 6: Assignment 6 - Question 2

Due on 2024-03-07, 23:59 IST

Simple Path Finding

Given an n×n binary Matrix A , where each entry is 0 or 1.
A has a unique path of 1's from  A[0][0] to A[n-1][n-1].
The path always goes Right (R) or Down (D).

Write a C Program.to print the directions of this path.

Note: You can assume that there is exactly one correct path.
All 1's in A are in this unique path, there are no dead ends.

Input
The first line contains the dimension of the matrix n. Assume n < 100.
The second line contains the contents of the matrix A, each row per line.

Output
The path of 1's in the Matrix.

Example 

Input

4
1 1 1 0 
0 0 1 1 
0 0 0 1 
0 0 0 1 


Output

RRDRDD

Explanation

The path of 1's from A[0][0] to A[3][3] is  

A[0][0]  Right --> A[0][1]  Right --> A[0][2]  Down --> A[1][2]  Right --> A[1][3]  Down --> A[2][3]  Down --> A[3][3].

Note: The code for reading inputs etc is given to you, complete the code of the function

void findPath(int matrix[100][100], int n, int x, int y, char* path, int pathIndex);

Your last recorded submission was on 2024-02-25, 12:34 IST

Select the Language for this assignment.      --                     C               

1

#include <stdio.h>

2

​

3

void findPath(int matrix[100][100], int n, int x, int y, char* path, int pathIndex) {

4

5

    // If the destination is reached, print the path and return

6

    if (x == n - 1 && y == n - 1) {

7

        path[pathIndex] = '\0'; // Null terminate the string

8

        printf("%s\n", path);

9

        return;

10

    }

11

​

12

    if ( matrix[x][y + 1] == 1) {

13

       path[pathIndex] = 'R';

14

       findPath(matrix, n, x, y+1, path, pathIndex+1);   

15

    }

16

    else if ( matrix[x+ 1][y] == 1 ){

17

       path[pathIndex] = 'D';

18

       findPath(matrix, n, x+1, y, path, pathIndex+1);

19

    }

20

}

21

​

22

int main() {

23

    int n;

24

    scanf("%d", &n);

25

26

    int matrix[100][100];

27

    char path[200]; // Assuming the path will not be longer than 2n-1 steps

28

​

29

    // Read the matrix

30

    for (int i = 0; i < n; i++) {

31

        for (int j = 0; j < n; j++) {

32

            scanf("%d", &matrix[i][j]);

33

        }

34

    }

35

36

    findPath(matrix, n, 0, 0, path, 0);

37

​

38

    return 0;

39

}

You may submit any number of times before the due date. The final submission will be considered for grading.

This assignment has Public Test cases. Please click on "Compile & Run" button to see the status of Public test cases. Assignment will be evaluated only after submitting using Submit button below. If you only save as or compile and run the Program , your assignment will not be graded and you will not see your score after the deadline.

   

Compilation : Passed
Public Test Cases: 1 / 1 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	4 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 1	RRDRDD	RRDRDD\n	Passed after ignoring Presentation Error

Private Test cases used for Evaluation	Status
Test Case 1	Passed
Test Case 2	Passed
Test Case 3	Passed

Week 6: Assignment 6 - Question 3

Due on 2024-03-07, 23:59 IST

Recursive Path Finding

Given an n×n binary Matrix A , where each entry is 0 or 1.
A has a unique path of 1's from  A[0][0] to A[n-1][n-1].
The path can go Right (R) Left (R)  Down (D) or Up (U).

Write a C Program.to print the directions of this path.

Note: You can assume that there is exactly one correct path.
All 1's in A need not be in this unique path, there can be dead ends.

Input
The first line contains the dimension of the matrix n. Assume n < 100.
The second line contains the contents of the matrix A, each row per line.

Output
The path of 1's in the Matrix.

Example 

Input

6

1 1 1 1 0 0

0 1 0 0 0 0

0 1 0 1 1 1

1 1 1 1 0 1

0 0 0 0 0 1

0 0 0 0 0 1

Output

RDDDRRURRDDD

Explanation

The correct path of 1's from A[0][0] to A[5][5] is 

A[0][0] Right  --> A[0][1] Down  --> A[1][1]  Down --> A[2][1]  Down --> A[3][1] Right --> A[3][2] Right  --> A[3][3] Up --> 

A[2][3] Right --> A[2][4] Right --> A[2][5] Down --> A[3][5] Down --> A[4][5] Down --> A[5][5].  

1 1 1 1 0 0

0 1 0 0 0 0

0 1 0 1 1 1

1 1 1 1 0 1

0 0 0 0 0 1

0 0 0 0 0 1

Note: The code for reading inputs etc is given to you, complete the code of the function

void findPath(int matrix[100][100], int n, int x, int y, char* path, int pathIndex);

Hint

Try all the paths LRDU one by one recursively [except the opposite of last direction taken].
If any of the recursive calls succeed, the function succeeds, return '1' immediately.
If all of the recursive calls fail, the function fails, return 0.

#include<stdio.h> int findPath(int matrix[100][100], int n, int x, int y, char* path, int pathIndex) { // If the destination is reached, print the path and return if (x == n - 1 && y == n - 1) { path[pathIndex] = '\0'; // Null terminate the string printf("%s", path); return 1; } int last = 'I'; if(pathIndex !=0) last = path[pathIndex - 1]; // Try moving Right if (last != 'L' && y + 1 < n && matrix[x][y + 1] == 1) { path[pathIndex] = 'R'; if(findPath(matrix, n, x, y+1, path, pathIndex+1) == 1) return 1; } // Try moving Down if (last != 'U' && x + 1 < n && matrix[x + 1][y] == 1) { path[pathIndex] = 'D'; if(findPath(matrix, n, x+1, y, path, pathIndex+1) == 1) return 1; } // Try moving Up if (last != 'D' && x - 1 >= 0 && matrix[x - 1][y] == 1) { path[pathIndex] = 'U'; if(findPath(matrix, n, x-1, y, path, pathIndex+1) == 1) return 1; } // Try moving Left if (last != 'R' && y - 1 >= 0 && matrix[x][y - 1] == 1) { path[pathIndex] = 'L'; if(findPath(matrix, n, x, y-1, path, pathIndex+1) == 1) return 1; } return 0; } int main() { int n; scanf("%d", &n); int matrix[100][100]; char path[1000]; // Read the matrix for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { scanf("%d", &matrix[i][j]); } } findPath(matrix, n, 0, 0, path, 0); return 0; }

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	6 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1	RDDDRRURRDDD	RDDDRRURRDDD	Passed
Test Case 2	4 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 1	RRDRDD	RRDRDD	Passed

*********

Week 7: Assignment 7 - Question 1

Due on 2024-03-14, 23:59 IST

CGPA calculation

You are the new tech administrator of a college. The college offers four courses listed below.

Course name       Course code          Credits
Science                1001                       10
Maths                   1002                       5
Literature              1003                       5
Philosophy            1004                      1
 
Each student has to take exactly three courses and she gets a grade from (0 - 10)
on the course, depending on her performance.

The CGPA of a student is calculated as  cgpa=∑3i=1creditsi∗gradei∑3i=1creditsi$����=\frac{\sum _{�=1}^3������{�}_{�}\ast ����{�}_{�}}{\sum _{�=1}^3������{�}_{�}}$.
That is for each course the student took, you multiply the grade obtained with the credits of the course.
Sum this value over each course, and divide it by the total credits of all courses the student took.

The previous administrator wrote an incomplete C program for a CGPA calculator, using structures.

Complete the C code for the program, by writing the code for the function 
float calculate_gpa(student s);

This function takes as an input the struct student variable of a student s, which contains all the information
about the courses the student s took and her grades in the courses.
You have to return the total cgpa of the student as the output.

Input
The first line contains the number of students n.
The next n lines contains the information on the students in the following order:
Name  CourseCode1 Marks1 CourseCode2 Marks2 CourseCode3 Marks3

Output
The names and CGPAs (to a single decimal point) of students, line by line in the following format:

Name CGPA

Note: Ignore the Passed after ignoring Presentation Error comment.

Example 

Input1

1

Akhil 1001 9 1002 10 1003 8

Output1
Akhil 9.0


Explanation

Akhil has scored
9/10 in 1001 (10 Credits)

10/10 in 1002 (5 Credits)

8/10 in 1003 (5 Credits)

So his total cgpa is  9∗10+10∗5+8∗510+5+5=18020=9.0$\frac{9\ast 10+10\ast 5+8\ast 5}{10+5+5}=\frac{180}{20}=9.0$.

Input2

4

Akash  1001 10 1002 5 1003 8

Akshat 1004 5 1001 10 1002 10

Amey   1002 0 1004 5 1001 10

Anuj     1002 5 1004 10 1003 0

Output2

Akash 8.2

Akshat 9.7

Amey 6.6

Anuj 3.2

Explanation

Akash has total cgpa 10∗10+5∗5+8∗510+5+5=16520=8.25$\frac{10\ast 10+5\ast 5+8\ast 5}{10+5+5}=\frac{165}{20}=8.25$, which is 8.2 after rounding off to one decimal place.
Similarly, the rest follows.

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Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	1 Akhil 1001 9 1002 10 1003 8	Akhil 9.0	Akhil 9.0\n	Passed after ignoring Presentation Error
Test Case 2	4 Akash 1001 10 1002 5 1003 8 Akshat 1004 5 1001 10 1002 10 Amey 1002 0 1004 5 1001 10 Anuj 1002 5 1004 10 1003 0	Akash 8.2\n Akshat 9.7\n Amey 6.6\n Anuj 3.2	Akash 8.2\n Akshat 9.7\n Amey 6.6\n Anuj 3.2\n	Passed after ignoring Presentation Error

Week 7: Assignment 7 - Question 2

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#include <stdio.h> #include <stdlib.h> struct node { int id; int value; struct node * next; }; /* Complete the code for creating a new node return the pointer to newly created node */ struct node * create_node(int id, int val) { struct node *temp; temp = calloc(1,sizeof(struct node)); temp->id=id; temp->value=val; return temp; } /* Complete the code for adding node e to the front of the list Return a pointer to the starting of the new list */ struct node * append(struct node * list, struct node * e) { e->next = list; list = e; return list; } /* Complete the code for searching the node with id in the list Return a pointer to the node if found. Return NULL otherwise */ struct node * search(struct node * list, int id) { struct node *curr=list; while(curr!=NULL && curr->id!=id) curr=curr->next; return curr; } /* Complete the code for changing the value of node e with given id to val Do nothing if such a node is not found. */ void change_value(struct node * list, int id, int val) { struct node* s = search(list,id); if(s!=NULL) s->value=val; } int find_value(struct node * list, int id) { struct node * e = search(list, id); if (e != NULL) return e->value; return -1; } int main() { char op; int id, val; struct node * list = NULL; int flag = 1; do { scanf("%c ", & op); switch (op) { case 'A': scanf("%d %d", & id, & val); list = append(list, create_node(id, val)); break; case 'S': scanf("%d", & id); printf("%d %d\n", id, find_value(list, id)); break; case 'C': scanf("%d %d", & id, & val); change_value(list, id, val); break; case 'E': flag = 0; } } while (flag == 1); return 0; }

Week 8: Assignment 8 - Question 1

Due on 2024-03-21, 23:59 IST

Deletion from Doubly linked list

In this question, you have to write code to remove the first node in a doubly linked list containing a specified number. The code to create the linked list is already given.

The main() function calls a function removeData(head,tail,n) to remove the first node containing data n from the linked list.

Complete the code by writing the missing function. You can also write additional functions which you may need.

Input Format

You will be given a positive number N in the first line.

This will be followed by a line containing N integers.

This will be followed by an integer M.

Output format

The output will be the list after deleting M.

Note : Only delete the first occurrence of M in list. If M is not present in the list, no change should happen to the list. Maintain the remaining elements in the original order.

Note : You do not have to add any additional code to print the list - once you code the removeData() function, the code

will print the contents of the remaining list.

Note: Ignore the Passed after ignoring Presentation Error comment.

Example 

Sample Input1

4

1 2 3 4

3

Sample Output1

1,2,4

Explanation

3 was present in the list and was removed. 

The elements other than 3 are printed.

Sample Input2

5

1 2 5 3 4

6

Sample Output2

1,2,5,3,4

Explanation

6 was not present in the list so nothing was removed.

All the elements in the list are printed.

#include <stdio.h> #include <stdlib.h> struct DoublyLinkedList { int data; struct DoublyLinkedList * prev; struct DoublyLinkedList * next; }; /* -------------------------- Create a new node -------------------------- */ struct DoublyLinkedList * createNode(int n) { struct DoublyLinkedList * newNodeptr; newNodeptr = (struct DoublyLinkedList * ) malloc(sizeof(struct DoublyLinkedList)); newNodeptr -> data = n; newNodeptr -> prev = NULL; newNodeptr -> next = NULL; return newNodeptr; } /* -------------------------------- add a node at the end of a doubly linked list. Tailptr is the address of the pointer to the end of the current list. After adding the node, tail points to the new node inserted. -------------------------------- */ void appendNode(struct DoublyLinkedList ** tailptr, int n) { struct DoublyLinkedList * newNode; newNode = createNode(n); newNode -> prev = * tailptr; ( * tailptr) -> next = newNode; * tailptr = newNode; } void initializeList( struct DoublyLinkedList ** headptr, struct DoublyLinkedList ** tailptr, int n) { struct DoublyLinkedList * newNode; newNode = createNode(n); * headptr = newNode; * tailptr = newNode; return; } void printList( struct DoublyLinkedList * head, struct DoublyLinkedList * tail) { struct DoublyLinkedList * curr = head; while (curr != NULL) { if (curr -> next != NULL) printf("%d,", curr -> data); else printf("%d", curr -> data); curr = curr -> next; } return; } /* -------------------------------- remove the node that ptr points to. After removing the first node, we should reset head. After removing the last node, we should reset tail. -------------------------------- */ void removeNode( struct DoublyLinkedList ** headptr, struct DoublyLinkedList ** tailptr, struct DoublyLinkedList * ptr) { if (ptr == * headptr) { * headptr = ptr -> next; ( * headptr) -> prev = NULL; } else { if (ptr == * tailptr) { * tailptr = ptr -> prev; ( * tailptr) -> next = NULL; } else { ptr -> prev -> next = ptr -> next; ptr -> next -> prev = ptr -> prev; } } free(ptr); ptr = NULL; return; } void removeData( struct DoublyLinkedList ** headptr, struct DoublyLinkedList ** tailptr, int n) { struct DoublyLinkedList * curr; curr = * headptr; while (curr != NULL && curr -> data != n) { curr = curr -> next; } if (curr != NULL) { removeNode(headptr, tailptr, curr); } return; } int main() { int n; int i = 0; int m; struct DoublyLinkedList * head, * tail; scanf("%d", & n); if (n <= 0) { return 0; } scanf("%d", & m); initializeList( & head, & tail, m); for (i = 1; i < n; i++) { /* read the remaining elements */ scanf("%d", & m); appendNode( & tail, m); } scanf("%d", & n); removeData( & head, & tail, n); printList(head, tail); return 0; }

Compilation : Passed
Public Test Cases: 4 / 4 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	4 1 2 3 4 3	1,2,4	1,2,4	Passed
Test Case 2	5 1 2 5 3 4 6	1,2,5,3,4	1,2,5,3,4	Passed
Test Case 3	4 1 2 3 4 1	2,3,4	2,3,4	Passed
Test Case 4	4 1 2 3 4 4	1,2,3	1,2,3	Passed

Week 8: Assignment 8 - Question 2

Due on 2024-03-21, 23:59 IST

Priority Queue using Linked List 

In this question, a  linked list is partially implemented where each element in the linked list is a structure of the following format:

struct node{

int id;

int priority;
struct node *next;

}; 

The field priority is a positive integer, which denotes the priority of an element inside the list.
The higher the value of integer in this field, the higher priority.

You have to complete the C code for performing the following operations in the linked list:

1. Create and return a node e with given id and value val
   struct node * create_node(int id, int val);
   
   
2. Add an node e to the beginning of the list. Return the new list.
   struct node * append(struct node * list, struct node * e);
   
   
3. Search for a node e with id inside the list. Return a pointer to e  if found, else return NULL
   struct node * search(struct node * list, int id); 
   
   
4. Change the value of an element with given id (if found), in the list to the new value val. 
   void change_priority(struct node * list, int id, int val) ;
   
   
5. Extract the element in the list with maximum priority. Return pointer to new list.
   struct node* extract_max(struct node * list);

After extract_max, the element having the max priority is removed from the list. Extract max also prints the id of the removed element in the following format "Max: id".

Note: You can assume that the priority of each element in the list is unique.

Note: The code for manipulating the input as well as output is given to you. You only have to write code for the incomplete functions.

Input
-------
A set of lines, each lines containing a character representing the operation and its inputs.

The operation can be one of the following:

• A <id> <val>
  Add an node with id and val to the list, at the start of the list.
  
  
• C <id> <val>
  Change the priority field of the element with id to val.
  If an element with this id is not found, do nothing.
   
• S <id>
  If an element with the id is in the list print the id and the priority and a newline. 
  Else, print the id and -1 and a newline.
  
  
• M
  Extract the element in the list with maximum priority. Print the id of the element as "Max: id" 
  
  
• E
  End of input, exit from the program

Output
---------
The output of search queries and extract max operations.

Sample input
-----------------

A 1 10

A 2 20

S 2

A 3 30
S 3
M
S 3

C 2 30

S 2

E

Sample Output
---------------------

2 20

3 30

Max: 3

3 -1

2 30

Explanation
---------------

• The list is initially empty
  
  
• Add an element 1 with value 10
  
  list : (1,10) -> NULL
  
  
• Add an element 2 with value 20
  
  list : (2,20) -> (1,10) -> NULL
  
  
• Search for element with id 2, print 
  
  
  2 20
  
  
• Add an element 3 with value 30
  
  list : (3,30) - > (2,20) -> (1,10) -> NULL
  
  
• Extract Max prints
  
  Max: 3
  
  list : (2,20) -> (1,10) -> NULL
  
  
• Search for element with id 3, print 
  
  3 -1
  
  
• Change priority of 2 to 30
  
  list : (2,30) -> (1,10) -> NULL
  
  
• Search for element with id 2, print 
  
  
  2 30
  
  End of input

**8**

#include <stdio.h> #include <stdlib.h> struct node { int id; int priority; struct node * next; }; /* create_node : Create a new struct node e, with given id and priority as val. Return a pointer to e. */ struct node * create_node(int id, int val) { struct node * e; e = (struct node * ) malloc(sizeof(struct node)); e -> id = id; e -> priority = val; e -> next = NULL; return e; } /* append : Append the given node e, to the start of the list. Return the head of the list, which is now e. */ struct node * append(struct node * list, struct node * e) { if (list == NULL) list = e; else { e -> next = list; } return e; } /* search : Search for a node e in list, with given id. Return Return a pointer to e, if found. Return NULL otherwise. */ struct node * search(struct node * list, int id) { while (list != NULL) { if (list -> id == id) return list; list = list -> next; } return NULL; } /* change_priority : Change priority of element with given id (if found) to val. */ void change_priority(struct node * list, int id, int val) { struct node * e = search(list, id); if (e != NULL) e -> priority = val; } /* delete_node : Remove element e from list; Return pointer to new head of list; */ struct node * delete_node(struct node * list, struct node *e){ struct node * a = list; if(e == NULL) return list; if(e == list){ free(e); return e -> next; } while(a->next != NULL){ if(a -> next == e){ a -> next = e -> next; break; } a = a->next; } free(e); return list; } /* extract_max : Find the element e in list with maximum priority. Remove the element from the list. Return the start of new list. */ struct node* extract_max(struct node * list) { struct node * e = list; struct node * maxe = NULL; /* Complete the code find element with maximum priority in list. store it to the variable maxe. */ int max_value = e->priority; maxe = e; while (e != NULL) { if(e->priority > max_value) { max_value = e -> priority; maxe = e; } e = e -> next; } if(maxe != NULL){ printf("Max: %d\n", maxe -> id); }else{ printf("Max: -1\n"); } return delete_node(list,maxe); } /* find_priority : Searches for an element with given id. Returns its priority if found, -1 otherwise. */ int find_priority(struct node * list, int id) { struct node * e = search(list, id); if (e != NULL) return e -> priority; return -1; } int main() { char op; int id, val; struct node * list = NULL; int flag = 1; do { scanf("%c ", & op); switch (op) { case 'A': scanf("%d %d", & id, & val); list = append(list, create_node(id, val)); break; case 'S': scanf("%d", & id); printf("%d %d\n", id, find_priority(list, id)); break; case 'C': scanf("%d %d", & id, & val); change_priority(list, id, val); break; case 'M': list = extract_max(list); break; case 'E': flag = 0; } } while (flag == 1); return 0; }

Compilation : Passed
Public Test Cases: 2 / 2 Passed

Note: These tests may not be considered while scoring. Know more.

Public Test Cases	Input	Expected Output	Actual Output	Status
Test Case 1	A 1 10 A 2 20 S 2 A 3 30 S 3 M S 3 C 2 30 S 2 E	2 20\n 3 30\n Max: 3\n 3 -1\n 2 30	2 20\n 3 30\n Max: 3\n 3 -1\n 2 30\n	Passed after ignoring Presentation Error
Test Case 2	A 1 10 A 2 20 A 3 30 S 3 S 4 C 2 40 M C 1 5 M A 4 50 S 4 A 5 40 S 5 S 1 E	3 30\n 4 -1\n Max: 2\n Max: 3\n 4 50\n 5 40\n 1 5	3 30\n 4 -1\n Max: 2\n Max: 3\n 4 50\n 5 40\n 1 5\n	Passed after ignoring Presentation Error