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# Find the subarray of size K with minimum XOR

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 `` ` `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;``     ` `    ``// 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

 ``

Output:

`16`

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

Related Topic: Subarrays, Subsequences, and Subsets in Array

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