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 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

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. If the count of 0s and 1s in the original string is not equal then print -1 else print the value of count after the traversal of the complete string.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output:

4

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



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Improved By : princiraj1992