# Length of the longest substring with equal 1s and 0s

Given a binary string. We need to find the length of longest balanced sub string. A sub string is balanced if it contains equal number of 0 and 1.

**Examples:**

Input : input = 110101010 Output : Length of longest balanced sub string = 8 Input : input = 0000 Output : Length of longest balanced sub string = 0

A **simple solution** is to use two nested loops to generate every substring. And a third loop to count number of 0s and 1s in current substring. Time complexity of this would be O(n^{3})

An **efficient solution** is to use hashing.

1) Traverse string and keep track of counts of 1s and 0s as count_1 and count_0 respectively.

2) See if current difference between two counts has appeared before (We use hashing to store all differences and first index where a difference appears). If yes, then substring from previous appearance and current index has same number of 0s and 1s.

Below is the implementation of above approach.

## C++

`// CPP for finding length of longest balanced ` `// substring ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns length of the longest substring ` `// with equal number of zeros and ones. ` `int` `stringLen(string str) ` `{ ` ` ` `// Create a map to store differences ` ` ` `// between counts of 1s and 0s. ` ` ` `map<` `int` `, ` `int` `> m; ` ` ` ` ` `// Initially difference is 0. ` ` ` `m[0] = -1; ` ` ` ` ` `int` `count_0 = 0, count_1 = 0; ` ` ` `int` `res = 0; ` ` ` `for` `(` `int` `i=0; i<str.size(); i++) ` ` ` `{ ` ` ` `// Keeping track of counts of ` ` ` `// 0s and 1s. ` ` ` `if` `(str[i] == ` `'0'` `) ` ` ` `count_0++; ` ` ` `else` ` ` `count_1++; ` ` ` ` ` `// If difference between current counts ` ` ` `// already exists, then substring between ` ` ` `// previous and current index has same ` ` ` `// no. of 0s and 1s. Update result if this ` ` ` `// substring is more than current result. ` ` ` `if` `(m.find(count_1 - count_0) != m.end()) ` ` ` `res = max(res, i - m[count_1 - count_0]); ` ` ` ` ` `// If current difference is seen first time. ` ` ` `else` ` ` `m[count_1 - count_0] = i; ` ` ` `} ` ` ` ` ` `return` `res; ` `} ` ` ` `// driver function ` `int` `main() ` `{ ` ` ` `string str = ` `"101001000"` `; ` ` ` `cout << ` `"Length of longest balanced"` ` ` `" sub string = "` `; ` ` ` `cout << stringLen(str); ` ` ` `return` `0; ` `}` |

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## Java

`// Java Code for finding the length of ` `// the longest balanced substring ` ` ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `public` `class` `MAX_LEN_0_1 { ` ` ` `public` `static` `void` `main(String args[])` `throws` `IOException ` ` ` `{ ` ` ` `String str = ` `"101001000"` `; ` ` ` ` ` `// Create a map to store differences ` ` ` `//between counts of 1s and 0s. ` ` ` `HashMap<Integer,Integer> map = ` `new` `HashMap<Integer,Integer>(); ` ` ` ` ` `// Initially difference is 0; ` ` ` `map. put(` `0` `, -` `1` `); ` ` ` `int` `res =` `0` `; ` ` ` `int` `count_0 = ` `0` `, count_1 = ` `0` `; ` ` ` `for` `(` `int` `i=` `0` `; i<str.length();i++) ` ` ` `{ ` ` ` `// Keep track of count of 0s and 1s ` ` ` `if` `(str.charAt(i)==` `'0'` `) ` ` ` `count_0++; ` ` ` `else` ` ` `count_1++; ` ` ` ` ` `// If difference between current counts ` ` ` `// already exists, then substring between ` ` ` `// previous and current index has same ` ` ` `// no. of 0s and 1s. Update result if this ` ` ` `// substring is more than current result. ` ` ` ` ` `if` `(map.containsKey(count_1-count_0)) ` ` ` `res = Math.max(res, (i - map.get(count_1-count_0))); ` ` ` ` ` `// If the current difference is seen first time. ` ` ` `else` ` ` `map.put(count_1-count_0,i); ` ` ` ` ` `} ` ` ` ` ` `System.out.println(` `"Length of longest balanced sub string = "` `+res); ` ` ` `} ` `} ` |

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## Python3

`# Python3 code for finding length of ` `# longest balanced substring ` ` ` `# Returns length of the longest substring ` `# with equal number of zeros and ones. ` `def` `stringLen( ` `str` `): ` ` ` ` ` `# Create a python dictionary to store ` ` ` `# differences between counts of 1s and 0s. ` ` ` `m ` `=` `dict` `() ` ` ` ` ` `# Initially difference is 0. ` ` ` `m[` `0` `] ` `=` `-` `1` ` ` ` ` `count_0 ` `=` `0` ` ` `count_1 ` `=` `0` ` ` `res ` `=` `0` ` ` `for` `i ` `in` `range` `(` `len` `(` `str` `)): ` ` ` ` ` `# Keeping track of counts of ` ` ` `# 0s and 1s. ` ` ` `if` `str` `[i] ` `=` `=` `'0'` `: ` ` ` `count_0 ` `+` `=` `1` ` ` `else` `: ` ` ` `count_1 ` `+` `=` `1` ` ` ` ` `# If difference between current ` ` ` `# counts already exists, then ` ` ` `# substring between previous and ` ` ` `# current index has same no. of ` ` ` `# 0s and 1s. Update result if ` ` ` `# this substring is more than ` ` ` `# current result. ` ` ` `if` `m.get(count_1 ` `-` `count_0): ` ` ` `res ` `=` `max` `(res, i ` `-` `m[count_1 ` `-` `count_0]) ` ` ` ` ` `# If current difference is ` ` ` `# seen first time. ` ` ` `else` `: ` ` ` `m[count_1 ` `-` `count_0] ` `=` `i ` ` ` `return` `res ` ` ` `# driver code ` `str` `=` `"101001000"` `print` `(` `"Length of longest balanced"` ` ` `" sub string = "` `,stringLen(` `str` `)) ` ` ` `# This code is contributed by "Sharad_Bhardwaj" ` |

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## C#

`// C# Code for finding the length of ` `// the longest balanced substring ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` `public` `static` `void` `Main(` `string` `[] args) ` `{ ` ` ` `string` `str = ` `"101001000"` `; ` ` ` ` ` `// Create a map to store differences ` ` ` `//between counts of 1s and 0s. ` ` ` `Dictionary<` `int` `, ` ` ` `int` `> map = ` `new` `Dictionary<` `int` `, ` ` ` `int` `>(); ` ` ` ` ` `// Initially difference is 0; ` ` ` `map[0] = -1; ` ` ` `int` `res = 0; ` ` ` `int` `count_0 = 0, count_1 = 0; ` ` ` `for` `(` `int` `i = 0; i < str.Length;i++) ` ` ` `{ ` ` ` `// Keep track of count of 0s and 1s ` ` ` `if` `(str[i] == ` `'0'` `) ` ` ` `{ ` ` ` `count_0++; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `count_1++; ` ` ` `} ` ` ` ` ` `// If difference between current counts ` ` ` `// already exists, then substring between ` ` ` `// previous and current index has same ` ` ` `// no. of 0s and 1s. Update result if this ` ` ` `// substring is more than current result. ` ` ` `if` `(map.ContainsKey(count_1 - count_0)) ` ` ` `{ ` ` ` `res = Math.Max(res, (i - map[count_1 - ` ` ` `count_0])); ` ` ` `} ` ` ` ` ` `// If the current difference is ` ` ` `// seen first time. ` ` ` `else` ` ` `{ ` ` ` `map[count_1 - count_0] = i; ` ` ` `} ` ` ` ` ` `} ` ` ` ` ` `Console.WriteLine(` `"Length of longest balanced"` `+ ` ` ` `" sub string = "` `+ res); ` `} ` `} ` ` ` `// This code is contributed by Shrikant13 ` |

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**Output:**

Length of longest balanced sub string = 6

**Time Complexity:** O(n)

Extended Problem : Largest subarray with equal number of 0s and 1s

This article is contributed by Shivam Pradhan (anuj_charm) and ASIPU PAWAN KUMAR. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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