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.

[sourcecode language=”CPP”]

// 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;

}

[/sourcecode]

**Output:**

0

**Time Complexity:** O (1)

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

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