Minimum number of characters to be removed to make a binary string alternate
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.
- If current and next are different then no need to perform deletion.
- 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 <bits/stdc++.h> 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. |
PHP
<?php // PHP 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. function countToMake0lternate( $s ) { $result = 0; for ( $i = 0; $i < ( strlen ( $s ) - 1); $i ++) // if two alternating characters // of string are same if ( $s [ $i ] == $s [ $i + 1]) // then need to // delete a character $result ++; return $result ; } // Driver code echo countToMake0lternate( "000111" ), "\n" ; echo countToMake0lternate( "11111" ), "\n" ; echo countToMake0lternate( "01010101" ) ; // This code is contributed by nitin mittal. ?> |
Javascript
<script> // Javascript 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. function countToMake0lternate(s) { let result = 0; for (let 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 document.write(countToMake0lternate( "000111" )+ "<br>" ); document.write(countToMake0lternate( "11111" )+ "<br>" ); document.write(countToMake0lternate( "01010101" )+ "<br>" ); // This code is contributed by rag2127 </script> |
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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