# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of the approach ` `#include ` `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

 `// 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

 `# 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#

 `// 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 `

Output:

```4
```

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

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Improved By : princiraj1992