Given two arrays A[] and **B[] **of equal length, the task is to find the Bitwise XOR of pairwise sum of the given two arrays.
Examples:
Input:_ A[] = {1, 2}, B[] = {3, 4} _
_Output: __2 _
Explanation:
_Sum of all possible pairs are {4(1 + 3), 5(1 + 4), 5(2 + 3), 6(2 + 4)} _
XOR of all the pair sums = 4 ^ 5 ^ 5 ^ 6 = 2
_Input: __A[] = {4, 6, 0, 0, 3, 3}, B[] = {0, 5, 6, 5, 0, 3} _
_Output: _8
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
**Naive Approach: **The simplest approach to solve the problem is to generate all possible pairs from the two given arrays and calculate their respective sums and update XOR with the sum of pairs. Finally, print the XOR obtained.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function to calculate the sum of
// XOR of the sum of every pair
**int** XorSum(``**int** A[], **int** B[], **int** N)
{
// Stores the XOR of sums
// of every pair
**int** ans = 0;
// Iterate to generate all possible pairs
**for** (``**int** i = 0; i < N; i++) {
**for** (``**int** j = 0; j < N; j++) {
// Update XOR
ans = ans ^ (A[i] + B[j]);
}
}
// Return the answer
**return** ans;
}
// Driver Code
**int** main()
{
**int** A[] = { 4, 6, 0, 0, 3, 3 };
**int** B[] = { 0, 5, 6, 5, 0, 3 };
**int** N = **sizeof** A / **sizeof** A[0];
cout << XorSum(A, B, N) << endl;
**return** 0;
}
Python3
## Python3 program to implement
## the above approach
## Function to calculate the sum of
## XOR of the sum of every pair
**def** XorSum(A, B, N):
## Stores the XOR of sums
## of every pair
ans **=** 0
## Iterate to generate all
## possible pairs
**for** i **in** range``(N):
**for** j **in** range``(N):
## Update XOR
ans **=** ans ^ (A[i] **+** B[j])
## Return the answer
**return** ans
## Driver Code
**if** __name__ **==** "__main__"``:
A **=** [ 4``, 6``, 0``, 0``, 3``, 3 ]
B **=** [ 0``, 5``, 6``, 5``, 0``, 3 ]
N **=** len``(A)
print (XorSum(A, B, N))
## This code is contributed by chitranayal
Output:
8
Time Complexity:_ O(N2) _
Auxiliary Space:_ O(1)_
**Efficient Approach: **The above approach can be optimized using Bit Manipulation technique. Follow the steps below to solve the problem:
- Considering only the Kth bit, the task is to count the number of pairs (i, j) such that the Kth bit of (Ai + Bj) is set.
- If this number is odd, add** X = 2k** to the answer. We are only interested in the values of (ai, bj) in modulo** 2X**.
- Thus, replace ai with ai % (2X) and bj with bj % (2X), and assume that ai and bj < 2X.
- There are two cases when the kth bit of (ai + bj) is set:
- x ≤ ai + bj < 2x
- 3x ≤ ai + bj < 4x
- Hence, sort b[] in increasing order. For a fixed i, the set of j that satisfies X ≤ (ai +bj) < 2X forms an interval.
- Therefore, count the number of such j by Binary search. Similarly, handle the second case.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function to calculate the
// XOR of the sum of every pair
**int** XorSum(``**int** A[], **int** B[], **int** N)
{
// Stores the maximum bit
**const** **int** maxBit = 29;
**int** ans = 0;
// Look for all the k-th bit
**for** (``**int** k = 0; k < maxBit; k++) {
// Stores the modulo of
// elements B[] with (2^(k+1))
**int** C[N];
**for** (``**int** i = 0; i < N; i++) {
// Calculate modulo of
// array B[] with (2^(k+1))
C[i] = B[i] % (1 << (k + 1));
}
// Sort the array C[]
sort(C, C + N);
// Stores the total number
// whose k-th bit is set
**long** **long** count = 0;
**long** **long** l, r;
**for** (``**int** i = 0; i < N; i++) {
// Calculate and store the modulo
// of array A[] with (2^(k+1))
**int** x = A[i] % (1 << (k + 1));
// Lower bound to count the number
// of elements having k-th bit in
// the range (2^k - x, 2* 2^(k) - x)
l = lower_bound(C,
C + N,
(1 << k) - x)
- C;
r = lower_bound(C,
C + N,
(1 << k) * 2 - x)
- C;
// Add total number i.e (r - l)
// whose k-th bit is one
count += (r - l);
// Lower bound to count the number
// of elements having k-th bit in
// range (3 * 2^k - x, 4*2^(k) - x)
l = lower_bound(C,
C + N,
(1 << k) * 3 - x)
- C;
r = lower_bound(C,
C + N,
(1 << k) * 4 - x)
- C;
count += (r - l);
}
// If count is even, Xor of
// k-th bit becomes zero, no
// need to add to the answer.
// If count is odd, only then,
// add to the final answer
**if** (count & 1)
ans += (1 << k);
}
// Return answer
**return** ans;
}
// Driver code
**int** main()
{
**int** A[] = { 4, 6, 0, 0, 3, 3 };
**int** B[] = { 0, 5, 6, 5, 0, 3 };
**int** N = **sizeof** A / **sizeof** A[0];
// Function call
cout << XorSum(A, B, N) << endl;
**return** 0;
}
Output:
8
Time Complexity:_ O(NlogN) _
Auxiliary Space:_ O(N)_
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#arrays #bit magic #mathematical #sorting #binary search #bitwise-xor
