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Deletions of “01” or “10” in binary string to make it free from “01” or “10”
• Difficulty Level : Easy
• Last Updated : 18 Mar, 2019

Given a binary string str, the task is to find the count of deletion of the sub-string “01” or “10” from the string so that the given string is free from these sub-strings. Print the minimum number of deletions.

Examples:

Input: str = “11010”
Output: 2
The resultant string will be “1”

Input: str = “1000101”
Output: 3
Resultant string, str = “0”

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

Approach: We are deleting “01” and “10” and the binary string contains only characters ‘0’ and ‘1’. Therefore, minimum number of deletions will be equal to the minimum of the count of ‘0’ and ‘1’.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the count of deletions``// of sub-strings "01" or "10"``int` `substrDeletion(string str, ``int` `len)``{`` ` `    ``// To store the count of 0s and 1s``    ``int` `count0 = 0, count1 = 0;`` ` `    ``for` `(``int` `i = 0; i < len; i++) {``        ``if` `(str[i] == ``'0'``)``            ``count0++;``        ``else``            ``count1++;``    ``}`` ` `    ``return` `min(count0, count1);``}`` ` `// Driver code``int` `main()``{``    ``string str = ``"010"``;``    ``int` `len = str.length();``    ``cout << substrDeletion(str, len);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``     ` `// Function to return the count of deletions``// of sub-strings "01" or "10"``static` `int` `substrDeletion(String str, ``int` `len)``{`` ` `    ``// To store the count of 0s and 1s``    ``int` `count0 = ``0``, count1 = ``0``;`` ` `    ``for` `(``int` `i = ``0``; i < len; i++)``    ``{``        ``if` `(str.charAt(i) == ``'0'``)``            ``count0++;``        ``else``            ``count1++;``    ``}`` ` `    ``return` `Math.min(count0, count1);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String str = ``"010"``;``    ``int` `len = str.length();``    ``System.out.println(substrDeletion(str, len));``}``}`` ` `// This code is contributed by Code_Mech.`

## Python3

 `# Python3 implementation of the approach `` ` `# Function to return the count of ``# deletions of sub-strings "01" or "10" ``def` `substrDeletion(string, length) :`` ` `    ``# To store the count of 0s and 1s ``    ``count0 ``=` `0``;``    ``count1 ``=` `0``; `` ` `    ``for` `i ``in` `range``(length) :``        ``if` `(string[i] ``=``=` `'0'``) :``            ``count0 ``+``=` `1``; ``        ``else` `:``            ``count1 ``+``=` `1``; `` ` `    ``return` `min``(count0, count1); `` ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `:`` ` `    ``string ``=` `"010"``; ``    ``length ``=` `len``(string); ``     ` `    ``print``(substrDeletion(string, length)); `` ` `# This code is contributed by Ryuga`

## C#

 `// C# implementation of the approach``using` `System;`` ` `class` `GFG``{``     ` `// Function to return the count of deletions``// of sub-strings "01" or "10"``static` `int` `substrDeletion(``string` `str, ``int` `len)``{`` ` `    ``// To store the count of 0s and 1s``    ``int` `count0 = 0, count1 = 0;`` ` `    ``for` `(``int` `i = 0; i < len; i++)``    ``{``        ``if` `(str[i] == ``'0'``)``            ``count0++;``        ``else``            ``count1++;``    ``}`` ` `    ``return` `Math.Min(count0, count1);``}`` ` `// Driver code``public` `static` `void` `Main()``{``    ``string` `str = ``"010"``;``    ``int` `len = str.Length;``    ``Console.Write(substrDeletion(str, len));``}``}`` ` `// This code is contributed by Ita_c.`
Output:
```1
```

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