# Previous number same as 1’s complement in C++

C++Server Side ProgrammingProgramming

In this problem, we are given an integer n. Our task is to check weather the preceding number is equal to 1’s complement of the number.

Let’s take a few examples to understand our problem

Input: 12
Output: No
Explanation: (12)10 = (1100)2
Preceding number 11 = (1011)2
1’s complement of 12 = (0011)2
Input: 4
Output: Yes
Explanation: 4 = (100)2
Preceding number 3 = (011)2
1’s complement of 12 = (011)2

To solve this problem, we can use a simple approach which is by comparing the previous number and the 1’s complement of the number.

This approach is simple but consumes space and time. time complexity: O(n)

An effective solution could be using the general method that we seek to solve the problem. Here, only the number which are powers of 2 will satisfy the condition i.e. the previous number is equal to 1’s complement.

Program to show the implementation of our solution

## Example

Live Demo

#include <iostream>
using namespace std;
bool sameBits(unsigned long int n){
if ((n & (n - 1)) == 0)
return true;
return false;
}
int main(){
unsigned long int n = 64;
if(sameBits(n))
cout<<"Both are the same";
else
cout<<"Both aren't the same";
return 0;
}

## Output

Both are the same