#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)
Answer
5
Explanation
Each call compares the first two elements and then moves one place to the right. It adds 1 whenever the two neighbours differ, and the last single element adds 1 more (the size 1 case). So foo returns the number of unequal neighbour pairs plus 1, which is the number of runs of equal values.
In 0, 1, 2, 2, 2, 0, 0, 1, 1 there are 4 unequal pairs: (0, 1), (1, 2), (2, 0) and (0, 1). They split the array into the runs 0 | 1 | 2 2 2 | 0 0 | 1 1.
Value printed \(= 4 + 1 = 5\).
Answer: 5