Previous number same as 1’s complement

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++

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// An efficient C++ program to check if binary
// representations of n's predecessor and its
// 1's complement are same.
#include <bits/stdc++.h>
#define ull unsigned long long int
using namespace std;
  
// Returns true if binary representations of
// n's predecessor and it's 1's complement are same.
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;
}

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Java

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// An efficient java program to check if binary
// representations of n's predecessor and its
// 1's complement are same.
public class GFG {
      
    // Returns true if binary representations of
    // n's predecessor and it's 1's complement
    // are same.
    static boolean bit_check(int n)
    {
        if ((n & (n - 1)) == 0)
            return true;
        return false;
    }
   
    // Driver code    
    public static void main(String args[]) {
          
         int n = 14;
         if(bit_check(n))
            System.out.println ('1');
         else
            System.out.println('0');
           
    }
}
  
// This code is contributed by Sam007

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Python3

# An efficient Python 3 program to check
# if binary representations of n’s predecessor
# and its 1’s complement are same.

# Returns true if binary representations
# of n’s predecessor and it’s 1’s
# complement are same.
def bit_check(n):
if ((n & (n – 1)) == 0):
return True
return False

# Driver Code
if __name__ == ‘__main__’:
n = 14
if(bit_check(n)):
print(‘1’)
else:
print(‘0’)

# This code is contributed by
# Surendra_Gangwar

C#

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// An efficient C# program to check if binary
// representations of n's predecessor and its
// 1's complement are same.
using System;
using System.Collections.Generic;
  
class GFG {
    
    // Returns true if binary representations of
    // n's predecessor and it's 1's complement
    // are same.
    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');
    }
}
  
// This code is contributed by Sam007

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PHP

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<?php
// An efficient PHP program to check 
// if binary representations of n's 
// predecessor and its 1's complement
// are same.
  
// Returns true if binary 
// representations of n's 
// predecessor and it's 1's 
// complement are same.
function bit_check($n)
{
    if (($n & ($n - 1)) == 0)
        return 1;
    return 0;
}
  
    // Driver code
    $n = 14;
    echo bit_check($n);
  
// This code is contributed by Sam007.
?>

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Output:

0

Time Complexity: O (1)

Auxiliary Space : O (1) No extra space is getting used.



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