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

Given a binary string. We need to find the length of the longest balanced substring. A substring is balanced if it contains an 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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(n3)

Below is the implementation of the above approach:

## C++

 // C++ program to find the length of  // the longest balanced substring #include using namespace std;    // Function to check if a string contains // equal number of one and zeros or not bool isValid(string p) {     int n = p.length();     int c1 = 0, c0 = 0;            for(int i =0; i < n; i++)     {         if(p[i] == '0')             c0++;         if(p[i] == '1')             c1++;     }            return (c0 == c1) ? true : false; }    // Function to find the length of  // the longest balanced substring int longestSub(string s) {     int max_len = 0;     int n = s.length();            for(int i = 0; i < n; i++)     {         for(int j = i; j < n; j++)         {             if(isValid(s.substr(i, j - i + 1)) && max_len < j - i + 1)                 max_len = j - i + 1;         }                }     return max_len;        }    // Driver code int main() {     string s = "101001000";            // Function call     cout << longestSub(s);            return 0; }

## Python3

 # Python3 program to find the length of # the longest balanced substring    # Function to check if a contains # equal number of one and zeros or not def isValid(p):        n = len(p)     c1 = 0     c0 = 0        for i in range(n):         if (p[i] == '0'):             c0 += 1         if (p[i] == '1'):             c1 += 1        if (c0 == c1):         return True     else:         return False    # Function to find the length of # the longest balanced substring def longestSub(s):            max_len = 0     n = len(s)        for i in range(n):         for j in range(i, n):             if (isValid(s[i : j - i + 1]) and                      max_len < j - i + 1):                 max_len = j - i + 1        return max_len    # Driver code if __name__ == '__main__':            s = "101001000"        # Function call     print(longestSub(s))    # This code is contributed by mohit kumar 29

Output:

6

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++

 // C++ for finding length of longest balanced // substring #include 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 m;            // Initially difference is 0.     m[0] = -1;               int count_0 = 0, count_1 = 0;     int res = 0;     for (int i=0; i

## 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 map = new HashMap();            // Initially difference is 0;             map. put(0, -1);             int res =0;             int count_0 = 0, count_1 = 0;             for(int i=0; i

## 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"

## 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 map = new Dictionary();            // 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

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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