# Minimum operations required to convert a binary string to all 0s or all 1s

Given a binary string str, the task is to find the minimum number of operations required to make all the characters of the string same i.e. either the resultant string contains all 0s or all 1s. In a single operation, any block of consecutive 0s can be converted to a block of consecutive 1s of the same length and vice versa.

Examples:

Input: str = “000111”
Output: 1
In a single operation, either change all 0s to 1s
or change all 1s to 0s.

Input: str = “0011101010”
Output: 3
All the 1s can be converted to 0s in 3 operations.

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

Approach: The problem is to convert all the characters into a single one. Now since converting a whole consecutive group of characters counts as a single step. You can calculate the number of different groups as separated by each other due to the presence of other characters in between them. Now the number of steps would be simply the minimum of both the numbers. Hence, the answer will be the minimum of the count of consecutive blocks of 0s or the count of consecutive blocks of 1s.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of ` `// minimum operations required ` `int` `minOperations(string str, ``int` `n) ` `{ ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` ` `  `        ``// Increment count when consecutive ` `        ``// characters are different ` `        ``if` `(str[i] != str[i + 1]) ` `            ``count++; ` `    ``} ` ` `  `    ``// Answer is rounding off the ` `    ``// (count / 2) to lower ` `    ``return` `(count + 1) / 2; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"000111"``; ` `    ``int` `n = str.length(); ` ` `  `    ``cout << minOperations(str, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the count of ` `// minimum operations required ` `static` `int` `minOperations(String str, ``int` `n) ` `{ ` `    ``int` `count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++)  ` `    ``{ ` ` `  `        ``// Increment count when consecutive ` `        ``// characters are different ` `        ``if` `(str.charAt(i) != str.charAt(i + ``1``)) ` `            ``count++; ` `    ``} ` ` `  `    ``// Answer is rounding off the ` `    ``// (count / 2) to lower ` `    ``return` `(count + ``1``) / ``2``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``String str = ``"000111"``; ` `    ``int` `n = str.length(); ` ` `  `    ``System.out.println(minOperations(str, n)); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of ` `# minimum operations required ` `def` `minOperations(``str``, n): ` `    ``count ``=` `0` `    ``for` `i ``in` `range``(n ``-` `1``): ` ` `  `        ``# Increment count when consecutive ` `        ``# characters are different ` `        ``if` `(``str``[i] !``=` `str``[i ``+` `1``]): ` `            ``count ``+``=` `1` ` `  `    ``# Answer is rounding off the ` `    ``# (count / 2) to lower ` `    ``return` `(count ``+` `1``) ``/``/` `2` ` `  `# Driver code ` `str` `=` `"000111"` `n ``=` `len``(``str``) ` ` `  `print``(minOperations(``str``, n)) ` ` `  `# This code is contributed ` `# by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the count of ` `// minimum operations required ` `static` `int` `minOperations(``string` `str, ``int` `n) ` `{ ` `    ``int` `count = 0; ` `    ``for` `(``int` `i = 0; i < n - 1; i++)  ` `    ``{ ` ` `  `        ``// Increment count when consecutive ` `        ``// characters are different ` `        ``if` `(str[(i)] != str[(i + 1)]) ` `            ``count++; ` `    ``} ` ` `  `    ``// Answer is rounding off the ` `    ``// (count / 2) to lower ` `    ``return` `(count + 1) / 2; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``string` `str = ``"000111"``; ` `    ``int` `n = str.Length; ` ` `  `    ``Console.WriteLine(minOperations(str, n)); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```1
```

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