Check if all bits can be made same by flipping two consecutive bits
Given a binary string, the task is to find whether all the digits of the string can be made equal i.e either 0 or 1 by flipping two consecutive bits any number of times.
Examples:
Input: 01011
Output: YES
Explanation:
Flip 2nd and 3rd bit -> 00111,
again flipping 1’st and 2’nd bit -> 11111Input: 100011
Output: NO
Explanation:
No number of moves can ever
equalize all elements of the array.
Approach:
On careful observation, toggling of i’th and j’th bit can be done by toggling from i’th bit like (i, i+1), (i+1, i+2) …. (j-1, j) here every bit is toggling twice (if bit is toggle twice then its come to its initial value) except i and j then ultimately i’th and j’th bits toggle. Therefore, it can be said that it is only not possible to make all digits in binary string equal when the count of both 1 and 0 is odd.
Below is the implementation of the above approach:
C++
// C++ program for the // above approach #include <bits/stdc++.h> using namespace std; // Function to check if // Binary string can be // made equal string canMake(string& s) { int o = 0, z = 0; // Counting occurrence of // zero and one in binary // string for ( int i = 0; i < s.size(); i++) { if (s[i] - '0' == 1) o++; else z++; } // From above observation if (o % 2 == 1 && z % 2 == 1) return "NO" ; else return "YES" ; } // Driver code int main() { string s = "01011" ; cout << canMake(s) << '\n' ; return 0; } |
Java
// Java program for the above approach import java.io.*; public class GFG { // Function to check if // Binary string can be // made equal static String canMake(String s) { int o = 0 , z = 0 ; // Counting occurrence of // zero and one in binary // string for ( int i = 0 ; i < s.length(); i++) { if (s.charAt(i) - '0' == 1 ) o++; else z++; } // From above observation if (o % 2 == 1 && z % 2 == 1 ) return "NO" ; else return "YES" ; } // Driver code public static void main (String[] args) { String s = "01011" ; System.out.println(canMake(s)) ; } } // This code is contributed by AnkitRai01 |
Python3
# Python3 program for the above approach # Function to check if # Binary string can be # made equal def canMake(s) : o = 0 ; z = 0 ; # Counting occurrence of # zero and one in binary # string for i in range ( len (s)) : if ( ord (s[i]) - ord ( '0' ) = = 1 ) : o + = 1 ; else : z + = 1 ; # From above observation if (o % 2 = = 1 and z % 2 = = 1 ) : return "NO" ; else : return "YES" ; # Driver code if __name__ = = "__main__" : s = "01011" ; print (canMake(s)); # This code is contributed by AnkitRai01 |
C#
// C# program for the above approach using System; class GFG { // Function to check if // Binary string can be // made equal static string canMake( string s) { int o = 0, z = 0; // Counting occurrence of // zero and one in binary // string for ( int i = 0; i < s.Length; i++) { if (s[i] - '0' == 1) o++; else z++; } // From above observation if (o % 2 == 1 && z % 2 == 1) return "NO" ; else return "YES" ; } // Driver code public static void Main() { string s = "01011" ; Console.WriteLine(canMake(s)) ; } } // This code is contributed by AnkitRai01 |
Javascript
<script> // javascript program for the above approach // Function to check if // Binary string can be // made equal function canMake(s) { var o = 0, z = 0; // Counting occurrence of // zero and one in binary // string for (i = 0; i < s.length; i++) { if (s.charAt(i).charCodeAt(0) - '0' .charCodeAt(0) == 1) o++; else z++; } // From above observation if (o % 2 == 1 && z % 2 == 1) return "NO" ; else return "YES" ; } // Driver code var s = "01011" ; document.write(canMake(s)) ; // This code is contributed by Rajput-Ji </script> |
PHP
<?php // Function to check if // Binary string can be // made equal function canMake( $s ) { $o = 0; $z = 0; // Counting occurrence of // zero and one in binary // string for ( $i = 0; $i < strlen ( $s ); $i ++) { if ( $s [ $i ] - '0' == 1) { $o ++; } else { $z ++; } } // From above observation if ( $o % 2 == 1 && $z % 2 == 1) { return "NO" ; } else { return "YES" ; } } // Driver code $s = "01011" ; echo canMake( $s ) . "\n" ; ?> |
YES
Time Complexity: O(n), where n is the length of the given Binary number
Auxiliary space: O(1) since it is using constant space for variables