Given an integer X. The task is to find and return the array containing of powers of 2’s and the xor of the array is X.
Examples:
Input: X = 20
Output: 16 4Input: X = 15
Output: 1 2 4 8
Approach: The answer lies in the binary representation of the number X.
Since in the power of 2, there is only one set bit. If there are two distinct powers of 2’s present then the xor will be the addition of both the numbers.
Similarly, if xor of the whole array will be taken then it should be equal to X and that will be the binary representation of that number.
Since there is a distinct set bit in every power of 2’s, the xor and the sum of the elements of the array will be the same.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the required arrayvector<long> getArray(int n){ vector<long> ans; // Store the power of 2 long p2 = 1; // while n is greater than 0 while (n > 0) { // if there is 1 in binary // representation if (n & 1) ans.push_back(p2); // Divide n by 2 // Multiply p2 by 2 n >>= 1; p2 *= 2; } return ans;}// Driver codeint main(){ long n = 15; // Get the answer vector<long> ans = getArray(n); // Printing the array for(int i : ans) cout << i << " "; return 0;} |
Java
// Java implementation implementation// of the above approachimport java.util.*;class GFG {// Function to return the required arraystatic Vector<Long> getArray(int n){ Vector<Long> ans = new Vector<Long>(); // Store the power of 2 long p2 = 1; // while n is greater than 0 while (n > 0) { // if there is 1 in binary // representation if (n % 2 == 1) ans.add(p2); // Divide n by 2 // Multiply p2 by 2 n >>= 1; p2 *= 2; } return ans;}// Driver codepublic static void main(String[] args) { int n = 15; // Get the answer Vector<Long> ans = getArray(n); // Printing the array for(Long i : ans) System.out.print(i + " ");}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the above approach # Function to return the required array def getArray(n) : ans = []; # Store the power of 2 p2 = 1; # while n is greater than 0 while (n > 0) : # if there is 1 in binary # representation if (n & 1) : ans.append(p2); # Divide n by 2 # Multiply p2 by 2 n >>= 1; p2 *= 2; return ans; # Driver code if __name__ == "__main__" : n = 15; # Get the answer ans = getArray(n); # Printing the array for i in ans : print(i, end = " "); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {// Function to return the required arraystatic List<long> getArray(int n){ List<long> ans = new List<long>(); // Store the power of 2 long p2 = 1; // while n is greater than 0 while (n > 0) { // if there is 1 in binary // representation if (n % 2 == 1) ans.Add(p2); // Divide n by 2 // Multiply p2 by 2 n >>= 1; p2 *= 2; } return ans;}// Driver codepublic static void Main(String[] args) { int n = 15; // Get the answer List<long> ans = getArray(n); // Printing the array foreach(long i in ans) Console.Write(i + " ");}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript implementation of the above approach// Function to return the required arrayfunction getArray(n){ let ans = []; // Store the power of 2 let p2 = 1; // while n is greater than 0 while (n > 0) { // if there is 1 in binary // representation if (n & 1) ans.push(p2); // Divide n by 2 // Multiply p2 by 2 n >>= 1; p2 *= 2; } return ans;}// Driver code let n = 15; // Get the answer let ans = getArray(n); // Printing the array for(let i = 0; i < ans.length; i++) document.write(ans[i] + " ");</script> |
Output:
1 2 4 8
Time Complexity: O(log2n)
Auxiliary Space: O(log2n)
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