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

• Difficulty Level : Easy
• Last Updated : 26 Nov, 2021

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 string
Space Complexity: O(1)

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