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Previous number same as 1’s complement in C++
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
#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
Updated on: 03-Feb-2020
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