Find Index of 0 to be replaced with 1 to get longest continuous sequence of 1s in a binary array

Given an array of 0s and 1s, find the position of 0 to be replaced with 1 to get longest continuous sequence of 1s.Expected time complexity is O(n) and auxiliary space is O(1).
Example:

Input: 
   arr[] =  {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1}
Output:
  Index 9
Assuming array index starts from 0, replacing 0 with 1 at index 9 causes
the maximum continuous sequence of 1s.

Input: 
   arr[] =  {1, 1, 1, 1, 0}
Output:
  Index 4

We strongly recommend to minimize the browser and try this yourself first.

A Simple Solution is to traverse the array, for every 0, count the number of 1s on both sides of it. Keep track of maximum count for any 0. Finally return index of the 0 with maximum number of 1s around it. The time complexity of this solution is O(n2).

Using an Efficient Solution, the problem can solved in O(n) time. The idea is to keep track of three indexes, current index (curr), previous zero index (prev_zero) and previous to previous zero index (prev_prev_zero). Traverse the array, if current element is 0, calculate the difference between curr and prev_prev_zero (This difference minus one is the number of 1s around the prev_zero). If the difference between curr andprev_prev_zero is more than maximum so far, then update the maximum. Finally return index of the prev_zero with maximum difference.

Following is C++ implementation of the above algorithm.

// C++ program to find Index of 0 to be replaced with 1 to get // longest continuous sequence of 1s in a binary array #include<iostream> using namespace std;   // Returns index of 0 to be replaced with 1 to get longest // continuous sequence of 1s.  If there is no 0 in array, then // it returns -1. int maxOnesIndex(bool arr[], int n) {     int max_count = 0;  // for maximum number of 1 around a zero     int max_index;  // for storing result     int prev_zero = -1;  // index of previous zero     int prev_prev_zero = -1; // index of previous to previous zero       // Traverse the input array     for (int curr=0; curr<n; ++curr)     {         // If current element is 0, then calculate the difference         // between curr and prev_prev_zero         if (arr[curr] == 0)         {             // Update result if count of 1s around prev_zero is more             if (curr - prev_prev_zero > max_count)             {                 max_count = curr - prev_prev_zero;                 max_index = prev_zero;             }               // Update for next iteration             prev_prev_zero = prev_zero;             prev_zero = curr;         }     }       // Check for the last encountered zero     if (n-prev_prev_zero > max_count)        max_index = prev_zero;       return max_index; }   // Driver program int main() {     bool arr[] = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1};     int n = sizeof(arr)/sizeof(arr[0]);     cout << "Index of 0 to be replaced is "          << maxOnesIndex(arr, n);     return 0; }

Output:

Index of 0 to be replaced is 9

Time Complexity: O(n)
Auxiliary Space: O(1)