Length of the longest substring with equal 1s and 0s Last Updated : 22 Feb, 2023 Comments Improve Suggest changes Like Article Like Report 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 = 110101010Output : Length of longest balanced sub string = 8 Input : input = 0000Output : 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. Below is the implementation of the above approach: C++ // C++ program to find the length of // the longest balanced substring #include <bits/stdc++.h> 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; } // This code is contributed by Aditya Kumar (adityakumar129) Java // Java program to find the length of // the longest balanced substring import java.io.*; import java.util.*; class GFG { // Function to check if a string contains // equal number of one and zeros or not public static boolean isValid(String p) { int n = p.length(); int c1 = 0, c0 = 0; for (int i = 0; i < n; i++) { if(p.charAt(i) == '0') { c0++; } if(p.charAt(i) == '1') { c1++; } } if(c0 == c1) { return true; } else { return false; } } // Function to find the length of // the longest balanced substring public static 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.substring(i, j + 1)) && max_len < j - i + 1) { max_len = j - i + 1; } } } return max_len; } // Driver code public static void main (String[] args) { String s = "101001000"; // Function call System.out.println(longestSub(s)); } } // This code is contributed by avanitrachhadiya2155 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 C# // C# program to find the length of // the longest balanced substring using System; class GFG{ // Function to check if a string contains // equal number of one and zeros or not static 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++; } } if (c0 == c1) { return true; } else { return false; } } // Function to find the length of // the longest balanced substring public static 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.Substring(i, j - i + 1)) && max_len < j - i + 1) { max_len = j - i + 1; } } } return max_len; } // Driver code static public void Main() { string s = "101001000"; // Function call Console.WriteLine(longestSub(s)); } } // This code is contributed by rag2127 JavaScript <script> // Javascript program to find the length of // the longest balanced substring // Function to check if a string contains // equal number of one and zeros or not function isValid(p) { var n = p.length; var c1 = 0, c0 = 0; for(var 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 function longestSub(s) { var max_len = 0; var n = s.length; for(var i = 0; i < n; i++) { for(var 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 var s = "101001000"; // Function call document.write( longestSub(s)); </script> Output6 Time Complexity: O(N3)Auxiliary Space: O(1) An efficient solution is to use hashing. Traverse string and keep track of counts of 1s and 0s as count_1 and count_0 respectively. 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<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; } 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); } } 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)!=None: 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<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 JavaScript <script> // Javascript Code for finding the length of // the longest balanced substring let str = "101001000"; // Create a map to store differences // between counts of 1s and 0s. let map = new Map(); // Initially difference is 0; map.set(0, -1); let res =0; let count_0 = 0, count_1 = 0; for(let 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.has(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.set(count_1-count_0,i); } document.write( "Length of longest balanced sub string = "+res ); // This code is contributed by unknown2108 </script> OutputLength of longest balanced sub string = 6 Time Complexity: O(n)Auxiliary Space: O(n) Extended Problem: Largest subarray with equal number of 0s and 1s Comment More infoAdvertise with us Next Article Analysis of Algorithms S Shivam.Pradhan Improve Article Tags : Strings Hash DSA binary-string Practice Tags : HashStrings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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