Previous number same as 1’s complement
Last Updated :
17 May, 2021
Given a number check whether binary representation of its predecessor and its 1’s complement are same or not.
Examples:
Input : 14
Output : NO
Storing 14 as a 4 bit number, 14 (1110), its predecessor 13 (1101), its 1’s complement 1 (0001), 13 and 1 are not same in their binary representation and hence output is NO.
Input : 8
Output : YES
Storing 8 as a 4 bit number, 8 (1000), its predecessor 7 (0111), its 1’s complement 7 (0111), both its predecessor and its 1’s complement are 7 and hence output is YES.
Simple Approach: In this approach, we actually calculate the complement of the number.
1. Find binary representation of the number’s predecessor and it’s 1’s complement using simple decimal to binary representation technique.
2. Compare bit by bit to check whether they are equal or not.
3. If all bits are equal then print YES else print NO.
Time Complexity: O (log n), as binary representation of numbers is getting calculated.
Auxiliary Space: O (1), although auxiliary space is O (1) still some memory spaces are getting
used to store binary representation of the numbers.
Efficient Approach: Only numbers which are powers of 2 have binary representation of their predecessor and their 1’s complement as same.
1. Check whether a number is power of 2 or not.
2. If a number is power of 2 then print YES else print NO.
C++
#include <bits/stdc++.h>
#define ull unsigned long long int
using namespace std;
bool bit_check(ull n)
{
if ((n & (n - 1)) == 0)
return true ;
return false ;
}
int main()
{
ull n = 14;
cout << bit_check(n) << endl;
return 0;
}
|
Java
public class GFG {
static boolean bit_check( int n)
{
if ((n & (n - 1 )) == 0 )
return true ;
return false ;
}
public static void main(String args[]) {
int n = 14 ;
if (bit_check(n))
System.out.println ( '1' );
else
System.out.println( '0' );
}
}
|
Python3
def bit_check(n):
if ((n & (n - 1 )) = = 0 ):
return True
return False
if __name__ = = '__main__' :
n = 14
if (bit_check(n)):
print ( '1' )
else :
print ( '0' )
|
C#
using System;
using System.Collections.Generic;
class GFG {
static bool bit_check( int n)
{
if ((n & (n - 1)) == 0)
return true ;
return false ;
}
public static void Main()
{
int n = 14;
if (bit_check(n))
Console.WriteLine ( '1' );
else
Console.WriteLine ( '0' );
}
}
|
PHP
<?php
function bit_check( $n )
{
if (( $n & ( $n - 1)) == 0)
return 1;
return 0;
}
$n = 14;
echo bit_check( $n );
?>
|
Javascript
<script>
function bit_check(n)
{
if ((n & (n - 1)) == 0)
return true ;
return false ;
}
let n = 14;
if (bit_check(n))
document.write('1 ');
else
document.write(' 0');
</script>
|
Time Complexity: O (1)
Auxiliary Space : O (1) No extra space is getting used.
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