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

• Difficulty Level : Easy
• Last Updated : 30 Mar, 2023

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 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`

## 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++

 `#include` `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

## 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 code` `str` `=` `"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<``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));` `  ``}` `}`

## 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 code` `let 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).

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