Construct original array starting with K from an array of XOR of all elements except elements at same index
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:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void constructArray( int A[], int N,
int K)
{
int B[N];
int totalXOR = A[0] ^ K;
for ( int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
for ( int i = 0; i < N; i++) {
cout << B[i] << " " ;
}
}
int main()
{
int A[] = { 13, 14, 10, 6 }, K = 2;
int N = sizeof (A) / sizeof (A[0]);
constructArray(A, N, K);
return 0;
}
|
Java
class GFG{
static void constructArray( int A[], int N,
int K)
{
int B[] = new int [N];
int totalXOR = A[ 0 ] ^ K;
for ( int i = 0 ; i < N; i++)
B[i] = totalXOR ^ A[i];
for ( int i = 0 ; i < N; i++)
{
System.out.print(B[i] + " " );
}
}
public static void main(String[] args)
{
int A[] = { 13 , 14 , 10 , 6 }, K = 2 ;
int N = A.length;
constructArray(A, N, K);
}
}
|
Python3
def constructArray(A, N, K):
B = [ 0 ] * N;
totalXOR = A[ 0 ] ^ K;
for i in range (N):
B[i] = totalXOR ^ A[i];
for i in range (N):
print (B[i], end = " " );
if __name__ = = '__main__' :
A = [ 13 , 14 , 10 , 6 ];
K = 2 ;
N = len (A);
constructArray(A, N, K);
|
C#
using System;
using System.Collections;
class GFG {
static void constructArray( int [] A, int N,
int K)
{
int [] B = new int [N];
int totalXOR = A[0] ^ K;
for ( int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
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;
constructArray(A, N, K);
}
}
|
Javascript
<script>
function constructArray(A, N, K)
{
let B = new Array(N);
let totalXOR = A[0] ^ K;
for (let i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
for (let i = 0; i < N; i++) {
document.write(B[i] + " " );
}
}
let A = [ 13, 14, 10, 6 ], K = 2;
let N = A.length;
constructArray(A, N, K);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
11 Jun, 2021
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