Given an integer N, find the one’s complement of the integer.
Examples:
Input: N = 5
Output: 2
Explanation: Binary representation of 5 is “101”. Its one’s complement is “010” = 2.Input: N = 255
Output: 0Input: N = 26
Output: 5
Approach: Already the XOR-based approach and converting the number to binary number approaches are mentioned in Set-1 of the article. Here the number is converted by flipping bits and adding that power of 2 to the answer. Follow the steps mentioned below to implement it:
- Find the binary representation of N.
- For each bit, complement it and add the contribution of this bit to the final answer.
- Return the final answer
Below is the implementation of the above approach.
C++
// CPP program to find 1's complement of N.#include <bits/stdc++.h>using namespace std;// Find the 1's complement of Nint findComplement(int num){ int ans = 0; for (int i = 0; num > 0; i++) { ans += pow(2, i) * (!(num % 2)); num /= 2; } return ans;}// Driver codeint main(){ unsigned int N = 5; cout << findComplement(N); return 0;} |
Java
// Java code for the above approachimport java.util.*;public class GFG{ // Find the 1's complement of N static int findComplement(int num) { int ans = 0; for (int i = 0; num > 0; i++) { if(num % 2 == 0) { ans += (int)Math.pow(2, i) ; } num /= 2; } return ans; } // Driver code public static void main(String args[]) { Integer N = 5; System.out.println(findComplement((int)N)); }}// This code is contributed by Samim Hossain Mondal |
Python3
# python3 program to find 1's complement of N.# Find the 1's complement of Ndef findComplement(num): ans = 0 i = 0 while(True): if num == 0: break ans += pow(2, i) * (not (num % 2)) num //= 2 i += 1 return ans# Driver codeif __name__ == "__main__": N = 5 print(findComplement(N))# This code is contributed by rakeshsahni |
C#
// C# code for the above approachusing System;class GFG{ // Find the 1's complement of Nstatic int findComplement(int num){ int ans = 0; for (int i = 0; num > 0; i++) { if(num % 2 == 0) { ans += (int)Math.Pow(2, i) ; } num /= 2; } return ans;}// Driver codepublic static void Main(){ uint N = 5; Console.Write(findComplement((int)N));}}// This code is contributed by Samim Hossain Mondal |
Javascript
<script> // JavaScript code for the above approach // Find the 1's complement of N function findComplement(num) { let ans = 0; for (let i = 0; num > 0; i++) { ans += Math.pow(2, i) * (!(num % 2)); num = Math.floor(num / 2); } return ans; } // Driver code let N = 5; document.write(findComplement(N));// This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(log N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
