# Minimum number of characters to be removed to make a binary string alternate

• Difficulty Level : Easy
• Last Updated : 07 Jul, 2022

Given a binary string, the task is to find minimum number of characters to be removed from it so that it becomes alternate. A binary string is alternate if there are no two consecutive 0s or 1s.
Examples :

Input  : s = "000111"
Output : 4
We need to delete two 0s and
two 1s to make string alternate.

Input  : s = "0000"
Output : 3
We need to delete three characters
to make it alternate.

Input  :  s = "11111"
Output :  4

Input  : s = "01010101"
Output : 0

Input  : s = "101010"
Output : 0

This problem has below simple solution.
We traverse string from left to right and compare current character with next character.

1. If current and next are different then no need to perform deletion.
2. If current and next are same, we need to perform one delete operation to make them alternate.

Below is the implementation of above algorithm.

## C++

 // C++ program to find minimum number // of characters to be removed to make // a string alternate. #include using namespace std;   // Returns count of minimum characters to // be removed to make s alternate. void countToMake0lternate(const string& s) {     int result = 0;       for (int i = 0; i < (s.length() - 1); i++)           // if two alternating characters         // of string are same         if (s[i] == s[i + 1])             result++; // then need to     // delete a character       return result; }   // Driver code int main() {     cout << countToMake0lternate("000111") << endl;     cout << countToMake0lternate("11111") << endl;     cout << countToMake0lternate("01010101") << endl;     return 0; }

## Java

 // Java program to find minimum number // of characters to be removed to make // a string alternate. import java.io.*;   public class GFG {       // Returns count of minimum characters to     // be removed to make s alternate.     static int countToMake0lternate(String s)     {         int result = 0;           for (int i = 0; i < (s.length() - 1); i++)               // if two alternating characters             // of string are same             if (s.charAt(i) == s.charAt(i + 1))                 result++; // then need to         // delete a character           return result;     }       // Driver code     static public void main(String[] args)     {         System.out.println(countToMake0lternate("000111"));         System.out.println(countToMake0lternate("11111"));         System.out.println(countToMake0lternate("01010101"));     } }   // This code is contributed by vt_m.

## Python 3

 # Python 3 program to find minimum number # of characters to be removed to make # a string alternate.   # Returns count of minimum characters # to be removed to make s alternate. def countToMake0lternate(s):       result = 0       for i in range(len(s) - 1):           # if two alternating characters         # of string are same         if (s[i] == s[i + 1]):             result += 1 # then need to                         # delete a character       return result   # Driver code if __name__ == "__main__":           print(countToMake0lternate("000111"))     print(countToMake0lternate("11111"))     print(countToMake0lternate("01010101"))   # This code is contributed by ita_c

## C#

 // C# program to find minimum number // of characters to be removed to make // a string alternate. using System;   public class GFG {       // Returns count of minimum characters to     // be removed to make s alternate.     static int countToMake0lternate(string s)     {         int result = 0;           for (int i = 0; i < (s.Length - 1); i++)               // if two alternating characters             // of string are same             if (s[i] == s[i + 1])                 result++; // then need to         // delete a character           return result;     }       // Driver code     static public void Main()     {         Console.WriteLine(countToMake0lternate("000111"));         Console.WriteLine(countToMake0lternate("11111"));         Console.WriteLine(countToMake0lternate("01010101"));     } }   // This article is contributed by vt_m.



## Javascript



Output:

4
4
0

Time Complexity : O(n) where n is number of characters in input string.

Auxiliary Space: O(1)
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