# Maximum length of a substring required to be flipped repeatedly to make all characters of binary string equal to 0

Given a binary string **S**, the task is to find the maximum length of substrings required to be flipped repeatedly to make all the characters of a binary string equal to **‘0’**.

**Examples:**

Input:S = “010”Output:2Explanation:

Following are the order of flipping of substring of at least K for the value of K as 2:

- Flip the substring S[0, 2] of length 3(>= K) modifies the string S to “101”.
- Flip the substring S[0, 1] of length 2(>= K) modifies the string S to “011”.
- Flip the substring S[1, 2] of length 2(>= K) modifies the string S to “000”.
For the value of K as 2(which is the maximum possible value) all the characters of the string can be made 0. Therefore, print 2.

Input:S = “00001111”Output:4

**Approach:** The given problem can be solved by traversing the given string **S**, now if at any point the adjacent characters are not the same then flip one sub-string **LHS** or **RHS**. For that, take the maximum length of **LHS** and **RHS**. There can be multiple adjacent places where characters are not equal. For each pair of substrings, the maximum required will be different. Now to change all the characters to **‘0’** take the minimum among all those maximums. Follow the steps below to solve the problem:

- Initialize the variable, say
**ans**as**N**that stores the maximum possible value of**K**. - Iterate over the range
**[0, N – 1)**using the variable**i**and perform the following steps:- If the value of
**s[i]**and**s[i + 1]**are not the same, then update the value of**ans**to the minimum of**ans**or maximum of**(i + 1)**or**(N – i – 1)**.

- If the value of
- After performing the above steps, print the value of
**ans**as the result.

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 value of K` `// such that flipping substrings of size` `// at least K make all characters 0s` `int` `maximumK(string& S)` `{` ` ` `int` `N = S.length();` ` ` `// Stores the maximum value of K` ` ` `int` `ans = N;` ` ` `int` `flag = 0;` ` ` `// Traverse the given string S` ` ` `for` `(` `int` `i = 0; i < N - 1; i++) {` ` ` `// Store the minimum of the` ` ` `// maximum of LHS and RHS length` ` ` `if` `(S[i] != S[i + 1]) {` ` ` `// Flip performed` ` ` `flag = 1;` ` ` `ans = min(ans, max(i + 1,` ` ` `N - i - 1));` ` ` `}` ` ` `}` ` ` `// If no flips performed` ` ` `if` `(flag == 0)` ` ` `return` `0;` ` ` `// Return the possible value of K` ` ` `return` `ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `string S = ` `"010"` `;` ` ` `cout << maximumK(S);` ` ` `return` `0;` `}` |

## Java

`// Java code for above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the maximum value of K` `// such that flipping substrings of size` `// at least K make all characters 0s` `static` `int` `maximumK(String S)` `{` ` ` `int` `N = S.length();` ` ` `// Stores the maximum value of K` ` ` `int` `ans = N;` ` ` `int` `flag = ` `0` `;` ` ` `// Traverse the given string S` ` ` `for` `(` `int` `i = ` `0` `; i < N - ` `1` `; i++) {` ` ` `// Store the minimum of the` ` ` `// maximum of LHS and RHS length` ` ` `if` `(S.charAt(i) != S.charAt(i + ` `1` `)) {` ` ` `// Flip performed` ` ` `flag = ` `1` `;` ` ` `ans = Math.min(ans, Math.max(i + ` `1` `,` ` ` `N - i - ` `1` `));` ` ` `}` ` ` `}` ` ` `// If no flips performed` ` ` `if` `(flag == ` `0` `)` ` ` `return` `0` `;` ` ` `// Return the possible value of K` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `String S = ` `"010"` `;` ` ` `System.out.print(maximumK(S));` `}` `}` `// This code is contributed by target_2.` |

## Python3

`# Python 3 program for the above approach` `# Function to find the maximum value of K` `# such that flipping substrings of size` `# at least K make all characters 0s` `def` `maximumK(S):` ` ` `N ` `=` `len` `(S)` ` ` `# Stores the maximum value of K` ` ` `ans ` `=` `N` ` ` `flag ` `=` `0` ` ` `# Traverse the given string S` ` ` `for` `i ` `in` `range` `(N ` `-` `1` `):` ` ` ` ` `# Store the minimum of the` ` ` `# maximum of LHS and RHS length` ` ` `if` `(S[i] !` `=` `S[i ` `+` `1` `]):` ` ` `# Flip performed` ` ` `flag ` `=` `1` ` ` `ans ` `=` `min` `(ans, ` `max` `(i ` `+` `1` `,N ` `-` `i ` `-` `1` `))` ` ` `# If no flips performed` ` ` `if` `(flag ` `=` `=` `0` `):` ` ` `return` `0` ` ` `# Return the possible value of K` ` ` `return` `ans` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `S ` `=` `"010"` ` ` `print` `(maximumK(S))` ` ` ` ` `# This code is contributed by SURENDRA_GANGWAR.` |

## C#

`// C# code for the above approach` `using` `System;` `public` `class` `GFG {` ` ` `// Function to find the maximum value of K` ` ` `// such that flipping substrings of size` ` ` `// at least K make all characters 0s` ` ` `static` `int` `maximumK(String S)` ` ` `{` ` ` `int` `N = S.Length;` ` ` `// Stores the maximum value of K` ` ` `int` `ans = N;` ` ` `int` `flag = 0;` ` ` `// Traverse the given string S` ` ` `for` `(` `int` `i = 0; i < N - 1; i++) {` ` ` `// Store the minimum of the` ` ` `// maximum of LHS and RHS length` ` ` `if` `(S[i] != S[i + 1]) {` ` ` `// Flip performed` ` ` `flag = 1;` ` ` `ans = Math.Min(ans,` ` ` `Math.Max(i + 1, N - i - 1));` ` ` `}` ` ` `}` ` ` `// If no flips performed` ` ` `if` `(flag == 0)` ` ` `return` `0;` ` ` `// Return the possible value of K` ` ` `return` `ans;` ` ` `}` ` ` `// Driver Code` ` ` `static` `public` `void` `Main()` ` ` `{` ` ` `// Code` ` ` `string` `S = ` `"010"` `;` ` ` `Console.Write(maximumK(S));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to find the maximum value of K` `// such that flipping substrings of size` `// at least K make all characters 0s` `function` `maximumK(S)` `{` ` ` `let N = S.length;` ` ` `// Stores the maximum value of K` ` ` `let ans = N;` ` ` `let flag = 0;` ` ` `// Traverse the given string S` ` ` `for` `(let i = 0; i < N - 1; i++)` ` ` `{` ` ` ` ` `// Store the minimum of the` ` ` `// maximum of LHS and RHS length` ` ` `if` `(S[i] != S[i + 1])` ` ` `{` ` ` ` ` `// Flip performed` ` ` `flag = 1;` ` ` `ans = Math.min(ans, Math.max(i + 1, N - i - 1));` ` ` `}` ` ` `}` ` ` `// If no flips performed` ` ` `if` `(flag == 0) ` `return` `0;` ` ` `// Return the possible value of K` ` ` `return` `ans;` `}` `// Driver Code` `let S = ` `"010"` `;` `document.write(maximumK(S));` `// This code is contributed by gfgking.` `</script>` |

**Output:**

2

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