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(n^{2}).

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 *and *prev_prev_zero* is more than maximum so far, then update the maximum. Finally return index of the prev_zero with maximum difference.

Following are the implementations of the above algorithm.

## C++

`// 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; ` `} ` |

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## Java

`// Java program to find Index of 0 to be replaced with 1 to get ` `// longest continuous sequence of 1s in a binary array ` ` ` `import` `java.io.*; ` ` ` `class` `Binary ` `{ ` ` ` `// 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. ` ` ` `static` `int` `maxOnesIndex(` `int` `arr[], ` `int` `n) ` ` ` `{ ` ` ` `int` `max_count = ` `0` `; ` `// for maximum number of 1 around a zero ` ` ` `int` `max_index=` `0` `; ` `// 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 to test above function ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `arr[] = {` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `1` `, ` `1` `, ` `1` `, ` `0` `, ` `1` `, ` `1` `, ` `1` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println(` `"Index of 0 to be replaced is "` `+ ` ` ` `maxOnesIndex(arr, n)); ` ` ` `} ` `} ` `/* This code is contributed by Devesh Agrawal */` |

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## Python3

`# Python program to find Index ` `# of 0 to be replaced with 1 to get ` `# longest continuous sequence ` `# of 1s in a binary array ` ` ` `# 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. ` `def` `maxOnesIndex(arr,n): ` ` ` ` ` `# for maximum number of 1 around a zero ` ` ` `max_count ` `=` `0` ` ` ` ` `# for storing result ` ` ` `max_index ` `=` `0` ` ` ` ` `# index of previous zero ` ` ` `prev_zero ` `=` `-` `1` ` ` ` ` `# index of previous to previous zero ` ` ` `prev_prev_zero ` `=` `-` `1` ` ` ` ` `# Traverse the input array ` ` ` `for` `curr ` `in` `range` `(n): ` ` ` ` ` `# 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 ` ` ` `arr ` `=` `[` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `1` `, ` `1` `, ` `1` `, ` `0` `, ` `1` `, ` `1` `, ` `1` `] ` `n ` `=` `len` `(arr) ` ` ` `print` `(` `"Index of 0 to be replaced is "` `, ` ` ` `maxOnesIndex(arr, n)) ` ` ` `# This code is contributed ` `# by Anant Agarwal. ` |

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## C#

`// C# program to find Index of 0 to be replaced ` `// with 1 to get longest continuous sequence of ` `// 1s in a binary array ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// 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. ` ` ` `static` `int` `maxOnesIndex(` `int` `[]arr, ` `int` `n) ` ` ` `{ ` ` ` ` ` `// for maximum number of 1 around a zero ` ` ` `int` `max_count = 0; ` ` ` ` ` `// for storing result ` ` ` `int` `max_index = 0; ` ` ` ` ` `// index of previous zero ` ` ` `int` `prev_zero = -1; ` ` ` ` ` `// index of previous to previous zero ` ` ` `int` `prev_prev_zero = -1; ` ` ` ` ` `// 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 to test above function ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[]arr = {1, 1, 0, 0, 1, 0, 1, 1, 1, ` ` ` `0, 1, 1, 1}; ` ` ` `int` `n = arr.Length; ` ` ` `Console.Write(` `"Index of 0 to be replaced is "` ` ` `+ maxOnesIndex(arr, n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by nitin mittal. ` |

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## PHP

`<?php ` `// PHP program to find Index of 0 to be ` `// replaced with 1 to get longest continuous ` `// sequence of 1s in a binary array ` ` ` `// 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. ` `function` `maxOnesIndex( ` `$arr` `, ` `$n` `) ` `{ ` ` ` `$max_count` `= 0; ` `// for maximum number of ` ` ` `// 1 around a zero ` ` ` `$max_index` `; ` `// for storing result ` ` ` `$prev_zero` `= -1; ` `// index of previous zero ` ` ` `$prev_prev_zero` `= -1; ` `// index of previous to ` ` ` `// previous zero ` ` ` ` ` `// Traverse the input array ` ` ` `for` `(` `$curr` `= 0; ` `$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 Code ` `$arr` `= ` `array` `(1, 1, 0, 0, 1, 0, 1, ` ` ` `1, 1, 0, 1, 1, 1); ` `$n` `= sizeof(` `$arr` `); ` `echo` `"Index of 0 to be replaced is "` `, ` ` ` `maxOnesIndex(` `$arr` `, ` `$n` `); ` ` ` `// This code is contributed by ajit ` `?> ` |

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**Output:**

Index of 0 to be replaced is 9

**Time Complexity: **O(n)

**Auxiliary Space:** O(1)

This article is contributed by **Ankur Singh**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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