GATE CS 2025 Set 1 Q.61: Pointers, Arrays and Strings

#include <stdio.h>
int foo(int S[],int size){
    if(size == 0) return 0;
    if(size == 1) return 1;
    if(S[0] != S[1]) return 1+foo(S+1,size-1);
    return foo(S+1,size-1);
}
int main(){
    int A[]={0,1,2,2,2,0,0,1,1};
    printf("%d",foo(A,9));
    return 0;
}

The value printed by the given C program is _______ . (Answer in integer)

Your answer

Official answer and solution

Official answer: 5
From: Official GATE 2025 answer key

Solution

Observe that foo(S, size) counts 1 for the final element plus 1 for each adjacent pair with S[i] != S[i+1], which equals the number of contiguous runs of identical elements.

For A = {0, 1, 2, 2, 2, 0, 0, 1, 1} of size = 9:

foo adds 1 whenever S[0] != S[1] (a transition between adjacent elements) and returns 1 at the base case size == 1.
In {0, 1, 2, 2, 2, 0, 0, 1, 1}, the transitions A[i] != A[i+1] occur at:
1. 0 != 1
2. 1 != 2
3. 2 != 0
4. 0 != 1
Adding 1 from the base case size == 1 gives $4 + 1 = 5$ (the 5 contiguous runs [0], [1], [2,2,2], [0,0], [1,1]).

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 61 of 65
Section: Computer Science and Information Technology

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