Given an array arr[] and integer K, the task is to find the minimum bitwise XOR sum of any subarray of size K in the given array.
Examples:
Input: arr[] = {3, 7, 90, 20, 10, 50, 40}, K = 3
Output: 16
Explanation:
The subarray {10, 50, 40} has the minimum XOR
Input: arr[] = {3, 7, 5, 20, -10, 0, 12}, K = 2
Output: 17
Explanation:
The subarray {5, 20} has the minimum XOR
Naive Approach: A Simple Solution is to consider every element as the beginning of subarray of size k and compute XOR of subarray starting with this element.
Time complexity: O(N * K)
Efficient Approach: The idea is to use the sliding window technique of size K and keep track of XOR sum of current K elements. To compute the XOR of the current window, perform XOR with the first element of the previous window to discard that element and with the current element to add this element into the window. Similarly, slide the windows to find the minimum XOR of the subarray of size K.
Below is the implementation of above approach:
C++
// C++ implementation to find the // subarray with minimum XOR #include <bits/stdc++.h> using namespace std; // Function to find the minimum // XOR of the subarray of size K void findMinXORSubarray(int arr[], int n, int k) { // K must be smaller // than or equal to n if (n < k) return; // Initialize beginning // index of result int res_index = 0; // Compute XOR sum of first // subarray of size K int curr_xor = 0; for (int i = 0; i < k; i++) curr_xor ^= arr[i]; // Initialize minimum XOR // sum as current xor int min_xor = curr_xor; // Traverse from (k+1)'th // element to n'th element for (int i = k; i < n; i++) { // XOR with current item // and first item of // previous subarray curr_xor ^= (arr[i] ^ arr[i - k]); // Update result if needed if (curr_xor < min_xor) { min_xor = curr_xor; res_index = (i - k + 1); } } cout << min_xor << "\n"; } // Driver Code int main() { int arr[] = { 3, 7, 90, 20, 10, 50, 40 }; int k = 3; // Subarray size int n = sizeof arr / sizeof arr[0]; // Function Call findMinXORSubarray(arr, n, k); return 0; } |
Java
// Java implementation to find the // subarray with minimum XOR class GFG{ // Function to find the minimum // XOR of the subarray of size K static void findMinXORSubarray(int arr[], int n, int k) { // K must be smaller // than or equal to n if (n < k) return; // Initialize beginning // index of result int res_index = 0; // Compute XOR sum of first // subarray of size K int curr_xor = 0; for(int i = 0; i < k; i++) curr_xor ^= arr[i]; // Initialize minimum XOR // sum as current xor int min_xor = curr_xor; // Traverse from (k+1)'th // element to n'th element for(int i = k; i < n; i++) { // XOR with current item // and first item of // previous subarray curr_xor ^= (arr[i] ^ arr[i - k]); // Update result if needed if (curr_xor < min_xor) { min_xor = curr_xor; res_index = (i - k + 1); } } System.out.print(min_xor + "\n"); } // Driver Code public static void main(String[] args) { int arr[] = { 3, 7, 90, 20, 10, 50, 40 }; // Subarray size int k = 3; int n = arr.length; // Function Call findMinXORSubarray(arr, n, k); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 implementation to find the # subarray with minimum XOR # Function to find the minimum # XOR of the subarray of size K def findMinXORSubarray(arr, n, k): # K must be smaller # than or equal to n if (n < k): return # Initialize beginning # index of result res_index = 0 # Compute XOR sum of first # subarray of size K curr_xor = 0 for i in range(0, k): curr_xor = curr_xor ^ arr[i] # Initialize minimum XOR # sum as current xor min_xor = curr_xor # Traverse from (k+1)'th # element to n'th element for i in range(k, n): # XOR with current item # and first item of # previous subarray curr_xor ^= (arr[i] ^ arr[i - k]) # Update result if needed if (curr_xor < min_xor): min_xor = curr_xor res_index = (i - k + 1) print(min_xor, end = '\n') # Driver Code arr = [ 3, 7, 90, 20, 10, 50, 40 ] # Subarray size k = 3 n = len(arr) # Function Call findMinXORSubarray(arr, n, k) # This code is contributed by PratikBasu |
C#
// C# implementation to find the // subarray with minimum XOR using System; class GFG{ // Function to find the minimum // XOR of the subarray of size K static void findMinXORSubarray(int []arr, int n, int k) { // K must be smaller // than or equal to n if (n < k) return; // Initialize beginning // index of result int res_index = 0; // Compute XOR sum of first // subarray of size K int curr_xor = 0; for(int i = 0; i < k; i++) curr_xor ^= arr[i]; // Initialize minimum XOR // sum as current xor int min_xor = curr_xor; // Traverse from (k+1)'th // element to n'th element for(int i = k; i < n; i++) { // XOR with current item // and first item of // previous subarray curr_xor ^= (arr[i] ^ arr[i - k]); // Update result if needed if (curr_xor < min_xor) { min_xor = curr_xor; res_index = (i - k + 1); } } Console.Write(min_xor + "\n"); } // Driver Code public static void Main(String[] args) { int []arr = { 3, 7, 90, 20, 10, 50, 40 }; // Subarray size int k = 3; int n = arr.Length; // Function Call findMinXORSubarray(arr, n, k); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript implementation to find the // subarray with minimum XOR // Function to find the minimum // XOR of the subarray of size K function findMinXORSubarray(arr, n, k) { // K must be smaller // than or equal to n if (n < k) return; // Initialize beginning // index of result let res_index = 0; // Compute XOR sum of first // subarray of size K let curr_xor = 0; for (let i = 0; i < k; i++) curr_xor ^= arr[i]; // Initialize minimum XOR // sum as current xor let min_xor = curr_xor; // Traverse from (k+1)'th // element to n'th element for (let i = k; i < n; i++) { // XOR with current item // and first item of // previous subarray curr_xor ^= (arr[i] ^ arr[i - k]); // Update result if needed if (curr_xor < min_xor) { min_xor = curr_xor; res_index = (i - k + 1); } } document.write(min_xor + "<br>"); } // Driver Code let arr = [ 3, 7, 90, 20, 10, 50, 40 ]; let k = 3; // Subarray size let n = arr.length; // Function Call findMinXORSubarray(arr, n, k); </script> |
Output:
16
Time Complexity: O(N)
Auxiliary Space: O(1)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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