XOR of all possible pairwise sum from two given Arrays

• Difficulty Level : Hard
• Last Updated : 14 Jun, 2021

Given two arrays A[] and B[] of equal length, the task is to find the Bitwise XOR of the pairwise sum of the given two arrays.

Examples:

Input: A[] = {1, 2}, B[] = {3, 4}
Output:
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

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 using namespace std; // Function to calculate the sum of// XOR of the sum of every pairint 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 Codeint main(){     int A[] = { 4, 6, 0, 0, 3, 3 };     int B[] = { 0, 5, 6, 5, 0, 3 };    int N = sizeof A / sizeof A;     cout << XorSum(A, B, N) << endl;     return 0;}

Java

 // Java program to implement// the above approachimport java.io.*; class GFG{ // Function to calculate the sum of// XOR of the sum of every pairstatic 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 Codepublic static void main (String[] args){    int A[] = { 4, 6, 0, 0, 3, 3 };    int B[] = { 0, 5, 6, 5, 0, 3 };         int N = A.length;         System.out.println(XorSum(A, B, N));}} // This code is contributed by AnkitRai01

Python3

 # Python3 program to implement# the above approach # Function to calculate the sum of# XOR of the sum of every pairdef 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 Codeif __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

C#

 // C# program to implement// the above approachusing System;class GFG{ // Function to calculate the sum of// XOR of the sum of every pairstatic 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 Codepublic static void Main(String[] args){    int []A = {4, 6, 0, 0, 3, 3};    int []B = {0, 5, 6, 5, 0, 3};    int N = A.Length;       Console.WriteLine(XorSum(A, B, N));}} // This code is contributed by Rajput-Ji

Javascript


Output:
8

Time Complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using the 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 using namespace std; // Function to calculate the// XOR of the sum of every pairint 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 codeint main(){    int A[] = { 4, 6, 0, 0, 3, 3 };    int B[] = { 0, 5, 6, 5, 0, 3 };    int N = sizeof A / sizeof A;     // Function call    cout << XorSum(A, B, N) << endl;     return 0;}

Java

 // Java Program to implement// the above approachimport java.util.*;class GFG{     // Lower boundstatic int lower_bound(int[] a, int low,                       int high, int element){    while(low < high)    {        int middle = low + (high - low) / 2;        if(element > a[middle])            low = middle + 1;        else            high = middle;    }    return low;} // Function to calculate the// XOR of the sum of every pairstatic int XorSum(int A[],                  int B[], int N){    // Stores the maximum bit    final 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 = new int[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[]    Arrays.sort(C);     // Stores the total number    // whose k-th bit is set    long count = 0;    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, 0, N,                       (1 << k) - x);         r = lower_bound(C, 0, N,                       (1 << k) *                        2 - x);         // 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, 0, N,                       (1 << k) *                        3 - x);        r = lower_bound(C, 0, N,                       (1 << k) *                        4 - x);         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) != 0)        ans += (1 << k);} // Return answerreturn ans;} // Driver codepublic static void main(String[] args){    int A[] = {4, 6, 0, 0, 3, 3};    int B[] = {0, 5, 6, 5, 0, 3};    int N = A.length;     // Function call    System.out.print(XorSum(A, B,                            N) + "\n");}} // This code is contributed by gauravrajput1

Python3

 # Python3 program to implement# the above approachfrom bisect import bisect, bisect_left, bisect_right # Function to calculate the# XOR of the sum of every pairdef XorSum(A, B, N):         # Stores the maximum bit    maxBit = 29     ans = 0     # Look for all the k-th bit    for k in range(maxBit):                 # Stores the modulo of        # elements B[] with (2^(k+1))        C =  * N                 for i in range(N):                         # Calculate modulo of            # array B[] with (2^(k+1))            C[i] = B[i] % (1 << (k + 1))         # Sort the array C[]        C = sorted(C)         # Stores the total number        # whose k-th bit is set        count = 0        l, r = 0, 0         for i in range(N):                         # Calculate and store the modulo            # of array A[] with (2^(k+1))            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 = bisect_left(C, (1 << k) - x)             r = bisect_left(C, (1 << k) * 2 - x)             # 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 = bisect_left(C, (1 << k) * 3 - x)            r = bisect_left(C, (1 << k) * 4 - x)                         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 codeif __name__ == '__main__':         A = [ 4, 6, 0, 0, 3, 3 ]    B = [ 0, 5, 6, 5, 0, 3 ]    N = len(A)     # Function call    print(XorSum(A, B, N)) # This code is contributed by mohit kumar 29

C#

 // C# program to implement// the above approachusing System; class GFG{ // Lower boundstatic int lower_bound(int[] a, int low,                       int high, int element){    while (low < high)    {        int middle = low + (high - low) / 2;        if (element > a[middle])            low = middle + 1;        else            high = middle;    }    return low;} // Function to calculate the// XOR of the sum of every pairstatic int XorSum(int[] A, int[] B, int N){         // Stores the maximum bit    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 = new int[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[]        Array.Sort(C);         // Stores the total number        // whose k-th bit is set        long count = 0;        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, 0, N,                            (1 << k) - x);             r = lower_bound(C, 0, N,                            (1 << k) * 2 - x);             // 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, 0, N,                            (1 << k) * 3 - x);            r = lower_bound(C, 0, N,                            (1 << k) * 4 - x);             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) != 0)            ans += (1 << k);    }     // Return answer    return ans;} // Driver codepublic static void Main(string[] args){    int[] A = { 4, 6, 0, 0, 3, 3 };    int[] B = { 0, 5, 6, 5, 0, 3 };    int N = A.Length;     // Function call    Console.Write(XorSum(A, B, N) + "\n");}} // This code is contributed by grand_master

Javascript


Output:
8

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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