Given an array A[] having n positive elements. The task to create another array B[] such as B[i] is XOR of all elements of array A[] except A[i].
Examples :
Input : A[] = {2, 1, 5, 9}
Output : B[] = {13, 14, 10, 6}
Input : A[] = {2, 1, 3, 6}
Output : B[] = {4, 7, 5, 0}
Naive Approach :
We can simple calculate B[i] as XOR of all elements of A[] except A[i], as
for (int i = 0; i < n; i++)
{
B[i] = 0;
for (int j = 0; j < n; j++)
if ( i != j)
B[i] ^= A[j];
}
Time complexity for this naive approach is O (n^2).
Auxiliary Space for this naive approach is O (n).
Optimized Approach :
First calculate XOR of all elements of array A[] say ‘xor’, and for each element of array A[] calculate A[i] = xor ^ A[i] .
int xor = 0;
for (int i = 0; i < n; i++)
xor ^= A[i];
for (int i = 0; i < n; i++)
A[i] = xor ^ A[i];
Time complexity for this approach is O (n).
Auxiliary Space for this approach is O (1).
C++
// C++ program to construct array from // XOR of elements of given array #include <bits/stdc++.h> using namespace std; // function to construct new array void constructXOR(int A[], int n) { // calculate xor of array int XOR = 0; for (int i = 0; i < n; i++) XOR ^= A[i]; // update array for (int i = 0; i < n; i++) A[i] = XOR ^ A[i]; } // Driver code int main() { int A[] = { 2, 4, 1, 3, 5}; int n = sizeof(A) / sizeof(A[0]); constructXOR(A, n); // print result for (int i = 0; i < n; i++) cout << A[i] << " "; return 0; } |
Java
// Java program to construct array from // XOR of elements of given array class GFG { // function to construct new array static void constructXOR(int A[], int n) { // calculate xor of array int XOR = 0; for (int i = 0; i < n; i++) XOR ^= A[i]; // update array for (int i = 0; i < n; i++) A[i] = XOR ^ A[i]; } // Driver code public static void main(String[] args) { int A[] = { 2, 4, 1, 3, 5}; int n = A.length; constructXOR(A, n); // print result for (int i = 0; i < n; i++) System.out.print(A[i] + " "); } } // This code is contributed by Anant Agarwal. |
Python3
# Python 3 program to construct # array from XOR of elements # of given array # function to construct new array def constructXOR(A, n): # calculate xor of array XOR = 0 for i in range(0, n): XOR ^= A[i] # update array for i in range(0, n): A[i] = XOR ^ A[i] # Driver code A = [ 2, 4, 1, 3, 5 ] n = len(A) constructXOR(A, n) # print result for i in range(0,n): print(A[i], end =" ") # This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to construct array from // XOR of elements of given array using System; class GFG { // function to construct new array static void constructXOR(int []A, int n) { // calculate xor of array int XOR = 0; for (int i = 0; i < n; i++) XOR ^= A[i]; // update array for (int i = 0; i < n; i++) A[i] = XOR ^ A[i]; } // Driver code public static void Main() { int []A = { 2, 4, 1, 3, 5}; int n = A.Length; constructXOR(A, n); // print result for (int i = 0; i < n; i++) Console.Write(A[i] + " "); } } // This code is contributed by nitin mittal |
PHP
<?php // Program to construct array from // XOR of elements of given array // function to construct new array function constructXOR(&$A, $n) { // calculate xor of array $XOR = 0; for ($i = 0; $i < $n; $i++) $XOR ^= $A[$i]; // update array for ($i = 0; $i < $n; $i++) $A[$i] = $XOR ^ $A[$i]; } // Driver code $A = array( 2, 4, 1, 3, 5); $n = sizeof($A); constructXOR($A, $n); // print result for ($i = 0; $i < $n; $i++) echo $A[$i] ." "; // This code is contributed // by ChitraNayal ?> |
Output:
3 5 0 2 4
Related Problem :
A Product Array Puzzle
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