XOR of Bitwise OR Pairs from Two Arrays
Last Updated : 31 Dec, 2023
Given two integer arrays A[] and B[] of size N and M respectively. Another array C is formed of size N*M, containing Bitwise OR of all possible pairs of elements of A with all elements of B i.e. A[i] | B[j] for all valid i and j. Find the Bitwise XOR of array C i.e. C [0] ^ C[1] ^ .......C[N*M].
Examples:
Input: N = 1, M = 3, A[] = {2}, B[] = {5, 0, 3}
Output: 6
Explanation: (2 OR 5) XOR (2 OR 0) XOR (2 OR 3) = 6.
Input: N = 2, M = 2, A[] = {1, 2}, B[] = {4, 10}
Output: 2
Explanation: Array C = [5, 11, 6, 10]. Xor is 2
Bruteforce Approach: The basic way to solve the problem:
The naive approach would be using nested loops to formulate all the possible pairs and find out the array C by evaluating the bitwise OR of every pair. Now calculate the bitwise XOR of the array C, to obtain the results.
Time complexity: O(N*M). where N and M are the lengths of the two arrays.
Efficient approach: To solve the problem follow the below intuition:
We need to check if the ith bit is set or not in the final answer. So for that we will maintain a count array, which will conatins the number of set bits in the array B at ith bit. Now we iterate on array A if the ith bit for a element in array A is a set bit then add M to the value if not add count[i] to the value. finally if the value is odd then the ith bit is set in the final answer.
Steps to solve the problem:
- Keep a track of count array of length 30. Where the ith index in the array i.e., count[i] denotes the number of set bits at ith position in the array B.
- Now iterate on each bit i, for all the elements in array A do the following:
- If ith bit is set add M i.e, length of array B to the variable value.
- else add count[i] to the variable value.
- If for a particular bit the the value obtained is odd then the bit is set in the final answer.
- Return answer after performing the above step on every bit.
Implementation of the above approach:
C++ // C++ code for the above approach: #include <iostream> using namespace std; int findORAndXOR(int A[], int B[], int N, int M) { // Count array to keep track of set bits int count[30] = { 0 }; // To store final answer int ans = 0; for (int i = 0; i < 30; i++) { for (int j = 0; j < M; j++) { if (B[j] & (1 << i)) count[i]++; } } for (int i = 0; i < 30; i++) { int value = 0; for (int j = 0; j < N; j++) { if (A[j] & (1 << i)) value += M; else value += count[i]; } // if value is odd. The current bit is // set in the final answer. if (value & 1) ans |= (1 << i); } return ans; } // Drivers code int main() { int N = 2, M = 2; int A[N] = { 1, 2 }; int B[M] = { 4, 10 }; // Function Call cout << findORAndXOR(A, B, N, M); return 0; } // Java code for the above approach import java.util.Arrays; class GFG { public static int findORAndXOR(int[] A, int[] B, int N, int M) { // Count array to keep track of set bits int[] count = new int[30]; // To store final answer int ans = 0; for (int i = 0; i < 30; i++) { for (int j = 0; j < M; j++) { if ((B[j] & (1 << i)) != 0) count[i]++; } } for (int i = 0; i < 30; i++) { int value = 0; for (int j = 0; j < N; j++) { if ((A[j] & (1 << i)) != 0) value += M; else value += count[i]; } // if value is odd. The current bit is // set in the final answer. if ((value & 1) != 0) ans |= (1 << i); } return ans; } // Driver code public static void main(String[] args) { int N = 2, M = 2; int[] A = { 1, 2 }; int[] B = { 4, 10 }; // Function Call System.out.println(findORAndXOR(A, B, N, M)); } } // This code is contributed by Abhinav Mahajan (abhinav_m22) def find_OR_and_XOR(A, B): # Count array to keep track of set bits count = [0] * 30 ans = 0 # Counting set bits in array B for i in range(30): for j in range(len(B)): if B[j] & (1 << i): count[i] += 1 # Calculating the final answer bit by bit for i in range(30): value = 0 # Computing the value for each bit position for j in range(len(A)): if A[j] & (1 << i): value += len(B) else: value += count[i] # Checking for odd value to set the bit in the final answer if value & 1: ans |= (1 << i) return ans # Driver code N = 2 M = 2 A = [1, 2] B = [4, 10] # Function Call and printing the result print(find_OR_and_XOR(A, B))
using System; class GFG { public static int FindORAndXOR(int[] A, int[] B, int N, int M) { // Count array to keep track of set bits int[] count = new int[30]; // To store final answer int ans = 0; for (int i = 0; i < 30; i++) { for (int j = 0; j < M; j++) { if ((B[j] & (1 << i)) != 0) count[i]++; } } for (int i = 0; i < 30; i++) { int value = 0; for (int j = 0; j < N; j++) { if ((A[j] & (1 << i)) != 0) value += M; else value += count[i]; } // if value is odd. The current bit is // set in the final answer. if ((value & 1) != 0) ans |= (1 << i); } return ans; } // Driver code public static void Main(string[] args) { int N = 2, M = 2; int[] A = { 1, 2 }; int[] B = { 4, 10 }; // Function Call Console.WriteLine(FindORAndXOR(A, B, N, M)); } } // Javascript code for the above approach: function findORAndXOR(A, B, N, M) { // Count array to keep track of set bits const count = Array(30).fill(0); // To store final answer let ans = 0; for (let i = 0; i < 30; i++) { for (let j = 0; j < M; j++) { if (B[j] & (1 << i)) count[i]++; } } for (let i = 0; i < 30; i++) { let value = 0; for (let j = 0; j < N; j++) { if (A[j] & (1 << i)) value += M; else value += count[i]; } // if value is odd. The current bit is // set in the final answer. if (value & 1) ans |= (1 << i); } return ans; } // Drivers code const N = 2, M = 2; const A = [1, 2]; const B = [4, 10]; // Function Call console.log(findORAndXOR(A, B, N, M)); // This code is contributed by ragul21 Time Complexity: O(30 * max(N, M)). Where N and M are the lengths of array A and B respectively.
Auxillary Space: O(30). Space used for count array.
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