Find One’s Complement of an Integer
Given an integer n, find the one’s complement of the integer.
Examples:
Input : n = 5 Output : 2 Input : n = 255 Output : 0 Input : n = 26 Output : 5
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Basic Approach :
The naïve approach to solve the problem would be to first convert the given number into its binary representation and then change every 1’s to 0 and 0’s to 1. After changing all 0’s and 1’s convert the binary representation to number.
Implementation of the above approach :
C++
// CPP program to find 1's complement of n.#include <bits/stdc++.h>using namespace std;unsigned int onesComplement(unsigned int n){ vector<int> v; // convert to binary representation while (n != 0) { v.push_back(n % 2); n = n / 2; } reverse(v.begin(), v.end()); // change 1's to 0 and 0's to 1 for (int i = 0; i < v.size(); i++) { if (v[i] == 0) v[i] = 1; else v[i] = 0; } // convert back to number representation int two = 1; for (int i = v.size() - 1; i >= 0; i--) { n = n + v[i] * two; two = two * 2; } return n;}int main(){ unsigned int n = 22; cout << onesComplement(n); return 0;} |
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An efficient approach to this problem is as follows:
1. Find the number of bits in the given integer
2. XOR the given integer with 2^number_of_bits-1
C++
// CPP program to find 1's complement of n.#include<bits/stdc++.h>using namespace std;unsigned int onesComplement(unsigned int n){ // Find number of bits in the given integer int number_of_bits = floor(log2(n))+1; // XOR the given integer with poe(2, // number_of_bits-1 and print the result return ((1 << number_of_bits) - 1) ^ n;}int main(){ unsigned int n = 22; cout << onesComplement(n); return 0;} |
Java
// Java program to find 1's complement of n.class GFG { static int onesComplement(int n) { // Find number of bits in the // given integer int number_of_bits = (int)(Math.floor(Math.log(n) / Math.log(2))) + 1; // XOR the given integer with poe(2, // number_of_bits-1 and print the result return ((1 << number_of_bits) - 1) ^ n; } // Driver code public static void main(String[] args) { int n = 22; System.out.print(onesComplement(n)); }}// This code is contributed by Anant Agarwal. |
Python3
# Python3 program to find# 1's complement of n.import mathdef onesComplement(n): # Find number of bits in # the given integer number_of_bits = (int)(math.floor(math.log(n) / math.log(2))) + 1; # XOR the given integer with poe(2, # number_of_bits-1 and print the result return ((1 << number_of_bits) - 1) ^ n;# Driver coden = 22print(onesComplement(n))# This code is contributed by Anant Agarwal. |
C#
// C# program to find 1's complement of n.using System;class GFG { static int onesComplement(int n) { // Find number of bits in the given integer int number_of_bits = (int)(Math.Floor( Math.Log(n) / Math.Log(2))) + 1; // XOR the given integer with poe(2, // number_of_bits-1 and print the result return ((1 << number_of_bits) - 1) ^ n; } //Driver code public static void Main () { int n = 22; Console.WriteLine(onesComplement(n)); }}// This code is contributed by Anant Agarwal. |
PHP
<?php// PHP program to find 1's// complement of n.function Log2($x){ return (log10($x) / log10(2));}function onesComplement($n){ // Find number of bits in// the given integer$number_of_bits = floor(log2($n)) + 1;// XOR the given integer with// number_of_bits-1 and print the resultreturn ((1 << $number_of_bits) - 1) ^ $n;}// Driver Code$n = 22;echo onesComplement($n);// This code is contributed by ihritik?> |
Javascript
<script>// Javascript program to find// 1's complement of n.// Function to check whether a// number is power of 2 or notfunction onesComplement(n) { // Find number of bits in the // given integer let number_of_bits = (Math.floor(Math.log(n) / Math.log(2))) + 1; // XOR the given integer with poe(2, // number_of_bits-1 and print the result return ((1 << number_of_bits) - 1) ^ n; }// driver program let n = 22; document.write(onesComplement(n)); </script> |
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-52W5vAk
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