# Maximum number of 0s that can be flipped such that Array has no adjacent 1s

Given a binary array **arr**, the task is to find the maximum number of 0s that can be flipped such that the array has no adjacent 1s, i.e. the array does not contain any two 1s at consecutive indices.**Examples:**

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

The 0 at index 2 can be replaced by 1.Input:arr[] = {1, 0, 0, 1}Output:0Explanation:

No 0 (zeroes) can be replaced by 1 such that no two consecutive indices have 1.

**Approach:**

- Iterate over the array and for every index which have 0, check if its adjacent two indices have 0 or not. For the last and first index of the array, check for the adjacent left and right index respectively.
- For every such index satisfying the above condition, increase the count.
- Print the final count at the end as the required answer

Below code is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Maximum number of 0s that` `// can be replaced by 1` `int` `canReplace(` `int` `array[], ` `int` `n)` `{` ` ` `int` `i = 0, count = 0;` ` ` `while` `(i < n)` ` ` `{` ` ` ` ` `// Check for three consecutive 0s` ` ` `if` `(array[i] == 0 &&` ` ` `(i == 0 || array[i - 1] == 0) &&` ` ` `(i == n - 1|| array[i + 1] == 0))` ` ` `{` ` ` `// Flip the bit` ` ` `array[i] = 1;` ` ` `// Increase the count` ` ` `count++;` ` ` `}` ` ` `i++;` ` ` `}` ` ` `return` `count;` `}` `// Driver's Code` `int` `main()` `{` ` ` `int` `array[5] = { 1, 0, 0, 0, 1 }; ` ` ` ` ` `cout << canReplace(array, 5);` `}` `// This code is contributed by spp____` |

## Java

`// Java program for the above approach` `public` `class` `geeks {` ` ` `// Maximum number of 0s that` ` ` `// can be replaced by 1` ` ` `public` `static` `int` `canReplace(` ` ` `int` `[] array)` ` ` `{` ` ` `int` `i = ` `0` `, count = ` `0` `;` ` ` `while` `(i < array.length) {` ` ` `// Check for three consecutive 0s` ` ` `if` `(array[i] == ` `0` ` ` `&& (i == ` `0` ` ` `|| array[i - ` `1` `] == ` `0` `)` ` ` `&& (i == array.length - ` `1` ` ` `|| array[i + ` `1` `] == ` `0` `)) {` ` ` `// Flip the bit` ` ` `array[i] = ` `1` `;` ` ` `// Increase the count` ` ` `count++;` ` ` `}` ` ` `i++;` ` ` `}` ` ` `return` `count;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `[] array = { ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `};` ` ` `System.out.println(canReplace(array));` ` ` `}` `}` |

## Python3

`# Python3 program for the above approach` `# Maximum number of 0s that` `# can be replaced by 1` `def` `canReplace(arr, n):` ` ` `i ` `=` `0` ` ` `count ` `=` `0` ` ` `while` `(i < n):` ` ` `# Check for three consecutive 0s` ` ` `if` `(arr[i] ` `=` `=` `0` `and` ` ` `(i ` `=` `=` `0` `or` `arr[i ` `-` `1` `] ` `=` `=` `0` `) ` `and` ` ` `(i ` `=` `=` `n ` `-` `1` `or` `arr[i ` `+` `1` `] ` `=` `=` `0` `)):` ` ` `# Flip the bit` ` ` `arr[i] ` `=` `1` ` ` `# Increase the count` ` ` `count ` `+` `=` `1` ` ` `i ` `+` `=` `1` ` ` `return` `count` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[ ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `]` ` ` ` ` `print` `(canReplace(arr, ` `5` `))` `# This code is contributed by himanshu77` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Maximum number of 0s that` `// can be replaced by 1` `public` `static` `int` `canReplace(` `int` `[] array)` `{` ` ` `int` `i = 0, count = 0;` ` ` `while` `(i < array.Length)` ` ` `{` ` ` `// Check for three consecutive 0s` ` ` `if` `(array[i] == 0 &&` ` ` `(i == 0 || array[i - 1] == 0) &&` ` ` `(i == array.Length - 1 || array[i + 1] == 0))` ` ` `{` ` ` ` ` `// Flip the bit` ` ` `array[i] = 1;` ` ` `// Increase the count` ` ` `count++;` ` ` `}` ` ` `i++;` ` ` `}` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `Main(String []args)` `{` ` ` `int` `[] array = { 1, 0, 0, 0, 1 };` ` ` ` ` `Console.WriteLine(canReplace(array));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` `// Javascript program for` `// the above approach` `// Maximum number of 0s that` `// can be replaced by 1` `function` `canReplace(array, n)` `{` ` ` `var` `i = 0, count = 0;` ` ` `while` `(i < n)` ` ` `{` ` ` ` ` `// Check for three consecutive 0s` ` ` `if` `(array[i] == 0 &&` ` ` `(i == 0 || array[i - 1] == 0) &&` ` ` `(i == n - 1|| array[i + 1] == 0))` ` ` `{` ` ` `// Flip the bit` ` ` `array[i] = 1;` ` ` `// Increase the count` ` ` `count++;` ` ` `}` ` ` `i++;` ` ` `}` ` ` `return` `count;` `}` `// Driver's Code` ` ` `array = [1, 0, 0, 0, 1] ` ` ` ` ` `document.write(canReplace(array, 5));` `</script>` |

**Output:**

1

**Time Complexity:** O(N)

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