Given an array A[] consisting of N integers and first element of the array B[] as K, the task is to construct the array B[] from A[] such that for any index i, A[i] is the Bitwise XOR of all the array elements of B[] except B[i].
Examples:
Input: A[] = {13, 14, 10, 6}, K = 2
Output: 2 1 5 9
Explanation:
For any index i, A[i] is the Bitwise XOR of all elements of B[] except B[i].
- B[1] ^ B[2] ^ B[3] = 1 ^ 5 ^ 9 = 13 = A[0]
- B[0] ^ B[2] ^ B[3] = 2 ^ 5 ^ 9 = 14 = A[1]
- B[0] ^ B[1] ^ B[3] = 2 ^ 1 ^ 9 = 10 = A[2]
- B[0] ^ B[1] ^ B[2] = 2 ^ 1 ^ 5 = 6 = A[3]
Input: A[] = {3, 5, 0, 2, 4}, K = 2
Output: 2 4 1 3 5
Approach: The idea is based on the observation that Bitwise XOR of the same value calculated even number of times is 0.
For any index i,
A[i] = B[0] ^ B[1] ^ … B[i-1] ^ B[i+1] ^ … B[n-1]
Therefore, XOR of all elements of B[], totalXor = B[0] ^ B[1] ^ … B[i – 1] ^ B[i] ^ B[i + 1] ^ … ^ B[N – 1].
Therefore, B[i] = totalXor ^ A[i]. (Since every element occurs twice except B[i])
Follow the below steps to solve the problem:
- Store the Bitwise XOR of all the elements present in the array B[] in a variable, say totalXOR, where totalXOR = A[0] ^ K.
- Traverse the given array A[] for each array element A[i], store the value of B[i] as totalXOR ^ A[i].
- After completing the above steps, print the element stored in the array B[].
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to construct an array // with each element equal to XOR // of all array elements except // the element at the same index void constructArray( int A[], int N,
int K)
{ // Original array
int B[N];
// Stores Bitwise XOR of array
int totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for ( int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for ( int i = 0; i < N; i++) {
cout << B[i] << " " ;
}
} // Driver Code int main()
{ int A[] = { 13, 14, 10, 6 }, K = 2;
int N = sizeof (A) / sizeof (A[0]);
// Function Call
constructArray(A, N, K);
return 0;
} |
// Java program for the above approach class GFG{
// Function to construct an array // with each element equal to XOR // of all array elements except // the element at the same index static void constructArray( int A[], int N,
int K)
{ // Original array
int B[] = new int [N];
// Stores Bitwise XOR of array
int totalXOR = A[ 0 ] ^ K;
// Calculate XOR of all array elements
for ( int i = 0 ; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for ( int i = 0 ; i < N; i++)
{
System.out.print(B[i] + " " );
}
} // Driver Code public static void main(String[] args)
{ int A[] = { 13 , 14 , 10 , 6 }, K = 2 ;
int N = A.length;
// Function Call
constructArray(A, N, K);
} } // This code is contributed by divyeshrabadiya07 |
# Python program for the above approach # Function to construct an array # with each element equal to XOR # of all array elements except # the element at the same index def constructArray(A, N, K):
# Original array
B = [ 0 ] * N;
# Stores Bitwise XOR of array
totalXOR = A[ 0 ] ^ K;
# Calculate XOR of all array elements
for i in range (N):
B[i] = totalXOR ^ A[i];
# Print the original array B
for i in range (N):
print (B[i], end = " " );
# Driver Code if __name__ = = '__main__' :
A = [ 13 , 14 , 10 , 6 ];
K = 2 ;
N = len (A);
# Function Call
constructArray(A, N, K);
# This code is contributed by Princi Singh |
// C# program for the above approach using System;
using System.Collections;
class GFG {
// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
static void constructArray( int [] A, int N,
int K)
{
// Original array
int [] B = new int [N];
// Stores Bitwise XOR of array
int totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for ( int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for ( int i = 0; i < N; i++)
{
Console.Write(B[i] + " " );
}
}
static void Main() {
int [] A = { 13, 14, 10, 6 };
int K = 2;
int N = A.Length;
// Function Call
constructArray(A, N, K);
}
} // This code is contributed by divyesh072019 |
<script> // JavaScript program for the above approach // Function to construct an array // with each element equal to XOR // of all array elements except // the element at the same index function constructArray(A, N, K)
{ // Original array
let B = new Array(N);
// Stores Bitwise XOR of array
let totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for (let i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for (let i = 0; i < N; i++) {
document.write(B[i] + " " );
}
} // Driver Code let A = [ 13, 14, 10, 6 ], K = 2;
let N = A.length;
// Function Call
constructArray(A, N, K);
// This code is contributed by Surbhi Tyagi. </script> |
2 1 5 9
Time Complexity: O(N)
Auxiliary Space: O(1)