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 -> 11111
Input: 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++
#include <bits/stdc++.h>
using namespace std;
string canMake(string& s)
{
int o = 0, z = 0;
for ( int i = 0; i < s.size(); i++) {
if (s[i] - '0' == 1)
o++;
else
z++;
}
if (o % 2 == 1 && z % 2 == 1)
return "NO" ;
else
return "YES" ;
}
int main()
{
string s = "01011" ;
cout << canMake(s) << '\n' ;
return 0;
}
|
Java
import java.io.*;
public class GFG
{
static String canMake(String s)
{
int o = 0 , z = 0 ;
for ( int i = 0 ; i < s.length(); i++)
{
if (s.charAt(i) - '0' == 1 )
o++;
else
z++;
}
if (o % 2 == 1 && z % 2 == 1 )
return "NO" ;
else
return "YES" ;
}
public static void main (String[] args)
{
String s = "01011" ;
System.out.println(canMake(s)) ;
}
}
|
Python3
def canMake(s) :
o = 0 ; z = 0 ;
for i in range ( len (s)) :
if ( ord (s[i]) - ord ( '0' ) = = 1 ) :
o + = 1 ;
else :
z + = 1 ;
if (o % 2 = = 1 and z % 2 = = 1 ) :
return "NO" ;
else :
return "YES" ;
if __name__ = = "__main__" :
s = "01011" ;
print (canMake(s));
|
C#
using System;
class GFG
{
static string canMake( string s)
{
int o = 0, z = 0;
for ( int i = 0; i < s.Length; i++)
{
if (s[i] - '0' == 1)
o++;
else
z++;
}
if (o % 2 == 1 && z % 2 == 1)
return "NO" ;
else
return "YES" ;
}
public static void Main()
{
string s = "01011" ;
Console.WriteLine(canMake(s)) ;
}
}
|
Javascript
<script>
function canMake(s)
{
var o = 0, z = 0;
for (i = 0; i < s.length; i++)
{
if (s.charAt(i).charCodeAt(0) - '0' .charCodeAt(0) == 1)
o++;
else
z++;
}
if (o % 2 == 1 && z % 2 == 1)
return "NO" ;
else
return "YES" ;
}
var s = "01011" ;
document.write(canMake(s)) ;
</script>
|
PHP
<?php
function canMake( $s ) {
$o = 0;
$z = 0;
for ( $i = 0; $i < strlen ( $s ); $i ++) {
if ( $s [ $i ] - '0' == 1) {
$o ++;
} else {
$z ++;
}
}
if ( $o % 2 == 1 && $z % 2 == 1) {
return "NO" ;
} else {
return "YES" ;
}
}
$s = "01011" ;
echo canMake( $s ) . "\n" ;
?>
|
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
Last Updated :
29 Apr, 2023
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