# Split the binary string into substrings with equal number of 0s and 1s

Given a binary string str of length N, the task is to find the maximum count of consecutive substrings str can be divided into such that all the substrings are balanced i.e. they have equal number of 0s and 1s. If it is not possible to split str satisfying the conditions then print -1.
Example:

Input: str = “0100110101”
Output:
The required substrings are “01”, “0011”, “01” and “01”.
Input: str = “0111100010”
Output:

Input: str = “0000000000”
Output: -1

Approach: Initialize count = 0 and traverse the string character by character and keep track of the number of 0s and 1s so far, whenever the count of 0s and 1s become equal increment the count. As in the given question, if it is not possible to split string then on that time count of 0s must not be equal to count of 1s then return -1 else print the value of count after the traversal of the complete string.
Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach#include using namespace std; // Function to return the count// of maximum substrings str// can be divided intoint maxSubStr(string str, int n){     // To store the count of 0s and 1s    int count0 = 0, count1 = 0;     // To store the count of maximum    // substrings str can be divided into    int cnt = 0;    for (int i = 0; i < n; i++) {        if (str[i] == '0') {            count0++;        }        else {            count1++;        }        if (count0 == count1) {            cnt++;        }    }     // It is not possible to    // split the string    if (count0!=count1) {        return -1;    }     return cnt;} // Driver codeint main(){    string str = "0100110101";    int n = str.length();     cout << maxSubStr(str, n);     return 0;}

## Java

 // Java implementation of the above approachclass GFG{ // Function to return the count// of maximum substrings str// can be divided intostatic int maxSubStr(String str, int n){     // To store the count of 0s and 1s    int count0 = 0, count1 = 0;     // To store the count of maximum    // substrings str can be divided into    int cnt = 0;    for (int i = 0; i < n; i++)    {        if (str.charAt(i) == '0')         {            count0++;        }        else        {            count1++;        }        if (count0 == count1)         {            cnt++;        }    }     // It is not possible to    // split the string    if (count0 != count1)     {        return -1;    }    return cnt;} // Driver codepublic static void main(String []args) {    String str = "0100110101";    int n = str.length();     System.out.println(maxSubStr(str, n));}} // This code is contributed by PrinciRaj1992

## Python3

 # Python3 implementation of the approach # Function to return the count # of maximum substrings str # can be divided intodef maxSubStr(str, n):         # To store the count of 0s and 1s    count0 = 0    count1 = 0         # To store the count of maximum     # substrings str can be divided into    cnt = 0         for i in range(n):        if str[i] == '0':            count0 += 1        else:            count1 += 1                     if count0 == count1:            cnt += 1 # It is not possible to     # split the string    if count0 != count1:        return -1                 return cnt # Driver codestr = "0100110101"n = len(str)print(maxSubStr(str, n))

## C#

 // C# implementation of the above approachusing System; class GFG{ // Function to return the count// of maximum substrings str// can be divided intostatic int maxSubStr(String str, int n){     // To store the count of 0s and 1s    int count0 = 0, count1 = 0;     // To store the count of maximum    // substrings str can be divided into    int cnt = 0;    for (int i = 0; i < n; i++)    {        if (str[i] == '0')         {            count0++;        }        else        {            count1++;        }        if (count0 == count1)         {            cnt++;        }    }     // It is not possible to    // split the string    if (count0 != count1)     {        return -1;    }    return cnt;} // Driver codepublic static void Main(String []args) {    String str = "0100110101";    int n = str.Length;     Console.WriteLine(maxSubStr(str, n));}} // This code is contributed by PrinciRaj1992

## Javascript



Output
4

Time complexity: O(N) where N is the length of the string
Space Complexity: O(1)

Another approach using Stack :
Approach: Similar to balanced parenthesis approach using stack, we keep inserting if top of stack matches with traversing character. we keep popping when its not matching with top of stack. Whenever stack is empty, it means we got a balanced substring. In this case, we increase answer variable. At last after complete traversal, we will check if stack is empty or not. If yes, it means everything is balanced out. If not, it means it’s not balanced.

Below is the implementation of the above approach:

## C++

 #includeusing namespace std;int maxSubStr(string str, int n){  //similar to balanced paranthesis approach  //we insert similar elements and pop when different element seen   //finally checking if stack will be empty or not at last  //if empty, it is balanced     int ans=0;     int i=0;     stacks;     s.push(str[i]);     i++;     while(i

## Java

 import java.util.*; class Main{  static int maxSubStr(String str, int n)  {    // similar to balanced paranthesis approach    // we insert similar elements and pop when different element seen     // finally checking if stack will be empty or not at last    // if empty, it is balanced    int ans = 0;    int i = 0;    Stack s = new Stack<>();    s.push(str.charAt(i));    i++;    while(i

## Python3

 def maxSubStr(s: str) -> int:    ans = 0    i = 0    stack = [s[i]]    i += 1    while i < len(s):        while i < len(s) and stack and i < len(s) and stack[-1] != s[i]:            stack.pop()            i += 1        if not stack:            ans += 1        while i < len(s) and (not stack or stack[-1] == s[i]):            stack.append(s[i])            i += 1    if not stack:        return ans    return -1 # Driver codestr = "0100110101"print(maxSubStr(str))  # This code is contributed by Aditya Sharma

## C#

 using System;using System.Collections.Generic; public class MaxSubStrSolution{  public static int MaxSubStr(string str, int n)  {     // similar to balanced paranthesis approach    // we insert similar elements and pop when different element seen     // finally checking if stack will be empty or not at last    // if empty, it is balanced    int ans = 0;    int i = 0;    Stack s = new Stack();    s.Push(str[i]);    i++;     while (i < str.Length)    {      while (i < str.Length && s.Count > 0 && i < str.Length && s.Peek() != str[i])      {        s.Pop();        i++;      }      if (s.Count == 0)      {        ans++;      }      while (i < str.Length && (s.Count == 0 || s.Peek() == str[i]))      {        s.Push(str[i]);        i++;      }    }     if (s.Count == 0)    {      return ans;    }     return -1;  }   // Driver code  public static void Main()  {    string str = "0100110101";    int n = str.Length;     Console.WriteLine(MaxSubStr(str, n));  }}

## Javascript

 const maxSubStr = (str, n) => {  //similar to balanced paranthesis approach  //we insert similar elements and pop when different element seen   //finally checking if stack will be empty or not at last  //if empty, it is balanced    let ans = 0;    let i = 0;    let s = [];    s.push(str[i]);    i++;    while(i < str.length){        while(i < str.length && s.length > 0 && s[s.length-1] !== str[i]){            s.pop();            i++;          }        if(s.length === 0)          ans++;        while(i < str.length && (s.length === 0 || s[s.length-1] === str[i])){            s.push(str[i]);            i++;        }    }    if(s.length === 0)      return ans;    return -1;}// Driver codelet str = "0100110101";let n = str.length; console.log(maxSubStr(str, n));

Output
4

Time complexity: O(n), where n is the length of the input string. This is because the code iterates over the string once.
Auxiliary Space: O(n), the above code is using a stack to store the elements of string, so over all complexity is O(n).

Previous
Next