# Minimum removal of consecutive similar characters required to empty a Binary String

Given a binary string S of length N, the task is to find the minimum number of removal of adjacent similar characters required to empty the given binary string.

Examples:

Input: S = “1100011“
Output: 2
Explanation:
Operation 1: Removal of all 0s modifies S to “1111“.
Operation 2: Removal of all remaining 1s makes S empty.
Therefore, the minimum number of operations required is 2.

Input: S = “0010100“
Output: 3
Explanation:
Operation 1: Removal of all 1s modifes S to “000100“.
Operation 2: Removal of all 1s modifies S = “00000“.
Operation 3: Removal of all remaining 0s makes S empty.
Therefore, the minimum number of operations required is 3.

Approach: The given problem can be solved using Greedy Approach. The idea is to delete the consecutive occurrences of the character with higher frequency. Follow the steps below to solve the problem:

• Traverse the given string S and generate a new string, say newString, by removing consecutive occurrences of the character with higher frequency.
• Finally, print (sizeof(newString) + 1)/2 as the required answer

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach`   `#include ` `using` `namespace` `std;`   `// Function to find minimum steps` `// to make the string empty` `int` `minSteps(string S)` `{` `    ``// Stores the modified string` `    ``string new_str;`   `    ``// Size of string` `    ``int` `N = S.length();`   `    ``int` `i = 0;`   `    ``while` `(i < N) {`   `        ``new_str += S[i];`   `        ``// Remving substring of same` `        ``// character from modified string` `        ``int` `j = i;` `        ``while` `(i < N && S[i] == S[j])` `            ``++i;` `    ``}`   `    ``// Print the minimum steps required` `    ``cout << ``ceil``((new_str.size() + 1) / 2.0);` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given string S` `    ``string S = ``"0010100"``;`   `    ``// Function Call` `    ``minSteps(S);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to find minimum steps` `// to make the String empty` `static` `void` `minSteps(String S)` `{` `    `  `    ``// Stores the modified String` `    ``String new_str = ``""``;`   `    ``// Size of String` `    ``int` `N = S.length();`   `    ``int` `i = ``0``;`   `    ``while` `(i < N) ` `    ``{` `        ``new_str += S.charAt(i);` `        `  `        ``// Remving subString of same` `        ``// character from modified String` `        ``int` `j = i;` `        ``while` `(i < N && S.charAt(i) == S.charAt(j))` `            ``++i;` `    ``}`   `    ``// Print the minimum steps required` `    ``System.out.print((``int``)Math.ceil(` `        ``(new_str.length() + ``1``) / ``2.0``));` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Given String S` `    ``String S = ``"0010100"``;`   `    ``// Function Call` `    ``minSteps(S);` `}` `}`   `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program for the above approach` `from` `math ``import` `ceil`   `# Function to find minimum steps` `# to make the empty` `def` `minSteps(S):` `    `  `    ``# Stores the modified string` `    ``new_str ``=` `""`   `    ``# Size of string` `    ``N ``=` `len``(S)`   `    ``i ``=` `0`   `    ``while` `(i < N):` `        ``new_str ``+``=` `S[i]`   `        ``# Removing substring of same character` `        ``# from modified string` `        ``j ``=` `i` `        ``while` `(i < N ``and` `S[i] ``=``=` `S[j]):` `            ``i ``+``=` `1`   `    ``# Print the minimum steps required` `    ``print``(ceil((``len``(new_str) ``+` `1``) ``/` `2``))`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Given S` `    ``S ``=` `"0010100"`   `    ``# Function Call` `    ``minSteps(S)`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach` `using` `System;`   `class` `GFG{`   `// Function to find minimum steps` `// to make the string empty` `static` `void` `minSteps(``string` `S)` `{` `    `  `    ``// Stores the modified string` `    ``string` `new_str = ``""``;` `    `  `    ``// Size of string` `    ``int` `N = S.Length;`   `    ``int` `i = 0;`   `    ``while` `(i < N) ` `    ``{` `        ``new_str += S[i];` `        `  `        ``// Remving substring of same` `        ``// character from modified string` `        ``int` `j = i;` `        `  `        ``while` `(i < N && S[i] == S[j])` `            ``++i;` `    ``}`   `    ``// Print the minimum steps required` `    ``Console.Write((``int``)Math.Ceiling(` `        ``(new_str.Length + 1) / 2.0));` `}`   `// Driver Code` `public` `static` `void` `Main()` `{` `    `  `    ``// Given string S` `    ``string` `S = ``"0010100"``;`   `    ``// Function Call` `    ``minSteps(S);` `}` `}`   `// This code is contributed by SURENDRA_GANGWAR`

Output:

```3

```

Time Complexity: O(N)
Auxiliary Space: O(1)

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