Minimum Lights To Activate

/*

Problem Description

There is a corridor in a Jail which is N units long. Given an array A of size N. The ith index of this array is 0 if the light at ith position is faulty otherwise it is 1.

All the lights are of specific power B which if is placed at position X, it can light the corridor from [ X-B+1, X+B-1].

Initially all lights are off.

Return the minimum number of lights to be turned ON to light the whole corridor or -1 if the whole corridor cannot be lighted.



Problem Constraints
1 <= N <= 1000

1 <= B <= 1000



Input Format
First argument is an integer array A where A[i] is either 0 or 1.

Second argument is an integer B.



Output Format
Return the minimum number of lights to be turned ON to light the whole corridor or -1 if the whole corridor cannot be lighted.


Example Input
Input 1:

A = [ 0, 0, 1, 1, 1, 0, 0, 1].
B = 3
Input 2:

A = [ 0, 0, 0, 1, 0].
B = 3


Example Output
Output 1:

2
Output 2:

-1

*/



int Solution::solve(vector<int> &A, int a) {
    int count = 0;
    int n = A.size();
    int i = 0;
    while(i < n){
        int right = min( i+a-1, n-1);// max coz should not go out of bound
        int left = max(i-a+1, 0);
        bool bulbFound = false;
        while(right >= left){
            if(A[right] == 1){
                bulbFound = true;
                break;
            }
            right--;
        }
        if(!bulbFound) return -1;
        count++;
        i = right + a;
        
    }

    return count;
}