GATE CS 2025 Set 1 Q51 — Recursion

The function foo(S, size) recursively traverses the array S from left to right. Let us analyze its behavior:

• Base Cases: If size == 0, it returns 0. If size == 1, it returns 1.
• Recursive Step: If the current element S[0] is not equal to the next element S[1], it adds 1 to the result of the recursive call on the remaining array. Otherwise, it simply returns the result of the recursive call without adding 1.

Thus, the function counts the number of adjacent mismatches ($S[i] \neq S[i+1]$) in the array and adds $1$ (from the base case when size == 1 is reached).

For the array A = {0, 1, 2, 2, 2, 0, 0, 1, 1} with size = 9, we check the adjacent pairs:

1. A[0] != A[1] ($0 \neq 1$) $\rightarrow$ True (adds 1)
2. A[1] != A[2] ($1 \neq 2$) $\rightarrow$ True (adds 1)
3. A[2] != A[3] ($2 \neq 2$) $\rightarrow$ False
4. A[3] != A[4] ($2 \neq 2$) $\rightarrow$ False
5. A[4] != A[5] ($2 \neq 0$) $\rightarrow$ True (adds 1)
6. A[5] != A[6] ($0 \neq 0$) $\rightarrow$ False
7. A[6] != A[7] ($0 \neq 1$) $\rightarrow$ True (adds 1)
8. A[7] != A[8] ($1 \neq 1$) $\rightarrow$ False

The recursion terminates at size == 1 (for the element A[8]), returning 1.

$$\text{Total Value} = 1 \text{ (from transitions)} + 1 \text{ (from transitions)} + 1 \text{ (from transitions)} + 1 \text{ (from transitions)} + 1 \text{ (base case)} = 5$$