#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)

The function foo(S, size) recursively scans adjacent elements of the array:

1. If size == 0, it returns $0$.

2. If size == 1, it returns $1$ (counting the last group/element).

3. If S[0] != S[1], it adds $1$ and recurses on the rest of the array (foo(S+1, size-1)).

4. If S[0] == S[1], it simply recurses on foo(S+1, size-1) without adding $1$.

Essentially, the function counts the number of contiguous blocks/runs of identical consecutive elements (or equivalently, the number of transitions between different consecutive elements plus $1$ for the final element).

Let us trace the execution for array $A = \{0, 1, 2, 2, 2, 0, 0, 1, 1\}$ with $\text{size} = 9$:

• foo(A, 9): $A[0] = 0, A[1] = 1 \implies 0 \neq 1 \implies 1 + \text{foo}(A+1, 8)$

• foo(A+1, 8): $A[1] = 1, A[2] = 2 \implies 1 \neq 2 \implies 1 + \text{foo}(A+2, 7)$

• foo(A+2, 7): $A[2] = 2, A[3] = 2 \implies 2 == 2 \implies \text{foo}(A+3, 6)$

• foo(A+3, 6): $A[3] = 2, A[4] = 2 \implies 2 == 2 \implies \text{foo}(A+4, 5)$

• foo(A+4, 5): $A[4] = 2, A[5] = 0 \implies 2 \neq 0 \implies 1 + \text{foo}(A+5, 4)$

• foo(A+5, 4): $A[5] = 0, A[6] = 0 \implies 0 == 0 \implies \text{foo}(A+6, 3)$

• foo(A+6, 3): $A[6] = 0, A[7] = 1 \implies 0 \neq 1 \implies 1 + \text{foo}(A+7, 2)$

• foo(A+7, 2): $A[7] = 1, A[8] = 1 \implies 1 == 1 \implies \text{foo}(A+8, 1)$

• foo(A+8, 1): Base case size == 1 reached, returns $1$.

Summing all returned values:

Thus, the printed value is 5.