# Maximize 0s to be flipped in given Binary array such that there are at least K 0s between two 1s

Given a binary array **arr[] **and an integer **K**, the task is to count the maximum number of** 0’s** that can be flipped to **1’s** such that there are at least **K** **0’s** between two **1’s**.

**Example:**

Input:arr[] = {0, 0, 1, 0, 0, 0}, K = 1Output:2Explanation:The 1st and the 5th index in the array arr[] can be flipped such that there is atleast 1 zero between any two 1’s. Therefore, the array after flipping is arr[] = {1, 0, 1, 0, 1, 0}.

Input:arr[] = {1, 0, 0, 0, 0, 0, 0, 0, 1, 0}, K = 2Output:1Explanation:The 4th index in the above array is the only valid index that can be flipped

**Approach: **The given problem can be solved by iterating through the array and finding the count of consecutive zeroes between the two 1’s. Suppose, the number of **0’s** between two **1’s** is **X**. Then, it can be observed that the number of **0’s** that can be flipped in between are **(X-K) / (K+1)**. Therefore, traverse the array and keep track of the number of consecutive **0’s** between two **1’s** similar to the algorithm discussed here and add the count of **0’s** that can be flipped into a variable **cnt**, which is the required answer.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the maximum number` `// of 1s that can be placed in array arr[]` `int` `maximumOnes(` `int` `arr[], ` `int` `N, ` `int` `K)` `{` ` ` `// Stores the count of 1's` ` ` `int` `cnt = 0;` ` ` `// Stores the last index of 1` ` ` `int` `last = -(K + 1);` ` ` `// Loop to iterate through the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If the current element is 1` ` ` `if` `(arr[i] == 1) {` ` ` `// Check if there are sufficient` ` ` `// 0's between consecutive 1's to` ` ` `// insert more 1's between them` ` ` `if` `(i - last - 1 >= 2 * (K - 1)) {` ` ` `cnt += (i - last - 1 - K) / (K + 1);` ` ` `}` ` ` `// Update the index of last 1` ` ` `last = i;` ` ` `}` ` ` `}` ` ` `// Condition to include the segment of` ` ` `// 0's in the last` ` ` `cnt += (N - last - 1) / (K + 1);` ` ` `// Return answer` ` ` `return` `cnt;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 0, 0, 0, 0, 0, 0, 0, 1, 0 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `int` `K = 2;` ` ` `cout << maximumOnes(arr, N, K);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function to find the maximum number` ` ` `// of 1s that can be placed in array arr[]` ` ` `static` `int` `maximumOnes(` `int` `arr[], ` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Stores the count of 1's` ` ` `int` `cnt = ` `0` `;` ` ` `// Stores the last index of 1` ` ` `int` `last = -(K + ` `1` `);` ` ` `// Loop to iterate through the array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) {` ` ` `// If the current element is 1` ` ` `if` `(arr[i] == ` `1` `) {` ` ` `// Check if there are sufficient` ` ` `// 0's between consecutive 1's to` ` ` `// insert more 1's between them` ` ` `if` `(i - last - ` `1` `>= ` `2` `* (K - ` `1` `)) {` ` ` `cnt += (i - last - ` `1` `- K) / (K + ` `1` `);` ` ` `}` ` ` `// Update the index of last 1` ` ` `last = i;` ` ` `}` ` ` `}` ` ` `// Condition to include the segment of` ` ` `// 0's in the last` ` ` `cnt += (N - last - ` `1` `) / (K + ` `1` `);` ` ` `// Return answer` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `};` ` ` `int` `N = arr.length;` ` ` `int` `K = ` `2` `;` ` ` `System.out.println(maximumOnes(arr, N, K));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## Python3

`# Python3 program for the above approach` `# Function to find the maximum number` `# of 1s that can be placed in array arr[]` `def` `maximumOnes(arr, N, K) :` ` ` `# Stores the count of 1's` ` ` `cnt ` `=` `0` `;` ` ` `# Stores the last index of 1` ` ` `last ` `=` `-` `(K ` `+` `1` `);` ` ` `# Loop to iterate through the array` ` ` `for` `i ` `in` `range` `(N) :` ` ` `# If the current element is 1` ` ` `if` `(arr[i] ` `=` `=` `1` `) :` ` ` `# Check if there are sufficient` ` ` `# 0's between consecutive 1's to` ` ` `# insert more 1's between them` ` ` `if` `(i ` `-` `last ` `-` `1` `>` `=` `2` `*` `(K ` `-` `1` `)) :` ` ` `cnt ` `+` `=` `(i ` `-` `last ` `-` `1` `-` `K) ` `/` `/` `(K ` `+` `1` `);` ` ` `# Update the index of last 1` ` ` `last ` `=` `i;` ` ` `# Condition to include the segment of` ` ` `# 0's in the last` ` ` `cnt ` `+` `=` `(N ` `-` `last ` `-` `1` `) ` `/` `/` `(K ` `+` `1` `);` ` ` `# Return answer` ` ` `return` `cnt;` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `1` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `];` ` ` `N ` `=` `len` `(arr);` ` ` `K ` `=` `2` `;` ` ` `print` `(maximumOnes(arr, N, K));` ` ` `# This code is contributed by AnkThon` |

## C#

`// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{` ` ` `// Function to find the maximum number` ` ` `// of 1s that can be placed in array arr[]` ` ` `static` `int` `maximumOnes(` `int` `[]arr, ` `int` `N, ` `int` `K)` ` ` `{` ` ` `// Stores the count of 1's` ` ` `int` `cnt = 0;` ` ` `// Stores the last index of 1` ` ` `int` `last = -(K + 1);` ` ` `// Loop to iterate through the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// If the current element is 1` ` ` `if` `(arr[i] == 1) {` ` ` `// Check if there are sufficient` ` ` `// 0's between consecutive 1's to` ` ` `// insert more 1's between them` ` ` `if` `(i - last - 1 >= 2 * (K - 1)) {` ` ` `cnt += (i - last - 1 - K) / (K + 1);` ` ` `}` ` ` `// Update the index of last 1` ` ` `last = i;` ` ` `}` ` ` `}` ` ` `// Condition to include the segment of` ` ` `// 0's in the last` ` ` `cnt += (N - last - 1) / (K + 1);` ` ` `// Return answer` ` ` `return` `cnt;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main(` `string` `[] args)` ` ` `{` ` ` `int` `[]arr = { 1, 0, 0, 0, 0, 0, 0, 0, 1, 0 };` ` ` `int` `N = arr.Length;` ` ` `int` `K = 2;` ` ` `Console.WriteLine(maximumOnes(arr, N, K));` ` ` `}` `}` `// This code is contributed by AnkThon` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to find the maximum number` `// of 1s that can be placed in array arr[]` `function` `maximumOnes(arr, N, K)` `{` ` ` `// Stores the count of 1's` ` ` `var` `cnt = 0;` ` ` `// Stores the last index of 1` ` ` `var` `last = -(K + 1);` ` ` `// Loop to iterate through the array` ` ` `for` `(` `var` `i = 0; i < N; i++) {` ` ` `// If the current element is 1` ` ` `if` `(arr[i] == 1) {` ` ` `// Check if there are sufficient` ` ` `// 0's between consecutive 1's to` ` ` `// insert more 1's between them` ` ` `if` `(i - last - 1 >= 2 * (K - 1)) {` ` ` `cnt += parseInt((i - last - 1 - K) / (K + 1));` ` ` `}` ` ` `// Update the index of last 1` ` ` `last = i;` ` ` `}` ` ` `}` ` ` `// Condition to include the segment of` ` ` `// 0's in the last` ` ` `cnt += parseInt((N - last - 1) / (K + 1));` ` ` `// Return answer` ` ` `return` `cnt;` `}` `// Driver Code` `var` `arr = [1, 0, 0, 0, 0, 0, 0, 0, 1, 0 ];` `var` `N = arr.length;` `var` `K = 2;` `document.write(maximumOnes(arr, N, K));` `// This code is contributed by rutvik_56.` `</script>` |

**Output**

1

**Time Complexity: **O(N)**Auxiliary Space: **O(1)