Skip to content
Related Articles

Related Articles

Improve Article

Find the subarray of size K with minimum XOR

  • Difficulty Level : Basic
  • Last Updated : 29 Apr, 2021

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 
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

 



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)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :