Array containing power of 2 whose XOR and Sum of elements equals X

• Last Updated : 17 Nov, 2021

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 4
Input: 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 using namespace std;   // Function to return the required array vector getArray(int n) {     vector 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 code int main() {     long n = 15;       // Get the answer     vector ans = getArray(n);       // Printing the array     for(int i : ans)         cout << i << " ";       return 0; }

Java

 // Java implementation implementation // of the above approach import java.util.*; class GFG {   // Function to return the required array static Vector getArray(int n) {     Vector ans = new Vector();       // 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 code public static void main(String[] args) {     int n = 15;       // Get the answer     Vector 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 approach using System; using System.Collections.Generic;   class GFG {   // Function to return the required array static List getArray(int n) {     List ans = new List();       // 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 code public static void Main(String[] args) {     int n = 15;       // Get the answer     List ans = getArray(n);       // Printing the array     foreach(long i in ans)         Console.Write(i + " "); } }   // This code is contributed by Princi Singh

Javascript



Output:

1 2 4 8

Time Complexity: O(n)

Auxiliary Space: O(n)

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