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

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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: 4 
The required substrings are "01", "0011", "01" and "01".
Input: str = "0111100010" 
Output: 3 

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++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count
// of maximum substrings str
// can be divided into
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 code
int main()
{
    string str = "0100110101";
    int n = str.length();

    cout << maxSubStr(str, n);

    return 0;
}

// Java implementation of the above approach
class GFG
{

// Function to return the count
// of maximum substrings str
// can be divided into
static 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 code
public 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 implementation of the approach

# Function to return the count 
# of maximum substrings str 
# can be divided into
def 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 code
str = "0100110101"
n = len(str)
print(maxSubStr(str, n))

// C# implementation of the above approach
using System;

class GFG
{

// Function to return the count
// of maximum substrings str
// can be divided into
static 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 code
public static void Main(String []args) 
{
    String str = "0100110101";
    int n = str.Length;

    Console.WriteLine(maxSubStr(str, n));
}
}

// This code is contributed by PrinciRaj1992

<script>

// JavaScript implementation of the approach

// Function to return the count
// of maximum substrings str
// can be divided into
function maxSubStr(str, n)
{

    // To store the count of 0s and 1s
    var count0 = 0, count1 = 0;

    // To store the count of maximum
    // substrings str can be divided into
    var cnt = 0;
    for (var 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 code
var str = "0100110101";
var n = str.length;
document.write( maxSubStr(str, n));

</script> 

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:

#include<bits/stdc++.h>
using 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;
     stack<int>s;
     s.push(str[i]);
     i++;
     while(i<str.size()){
        while(i<str.size()&&s.size()&&i<str.size()&&(s.top()!=str[i])){
            s.pop();
            i++;
        }
        if(s.empty()){
            ans++;
        }
        while((i<str.size())&&(s.empty()||s.top()==str[i])){
            s.push(str[i]);
            i++;
        }
     }
     if(s.empty())
     return ans;
     return -1;
     }
// Driver code
int main()
{
    string str = "0100110101";
    int n = str.length();
 
    cout << maxSubStr(str, n);
 
    return 0;
}

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<Character> s = new Stack<>();
    s.push(str.charAt(i));
    i++;
    while(i<str.length()){
      while(i<str.length() && !s.empty() && s.peek()!=str.charAt(i)){
        s.pop();
        i++;
      }
      if(s.empty()){
        ans++;
      }
      while(i<str.length() && (s.empty() || s.peek()==str.charAt(i))){
        s.push(str.charAt(i));
        i++;
      }
    }
    if(s.empty())
      return ans;
    return -1;
  }

  // Driver code
  public static void main(String[] args)
  {
    String str = "0100110101";
    int n = str.length();

    System.out.println(maxSubStr(str, n));

  }
}

// This code is contributed by anskalyan3

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 code
str = "0100110101"
print(maxSubStr(str)) 

# This code is contributed by Aditya Sharma

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<char> s = new Stack<char>();
    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));
  }
}

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 code
let str = "0100110101";
let n = str.length;

console.log(maxSubStr(str, n));

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).