Maximum XOR subarray of K distinct integer
Last Updated :
27 Apr, 2023
Given an array A consisting of N positive integers, the task is to calculate the maximum XOR of the subarray of size K consisting of all distinct integers. If such subarray is not present then Print "-1".
Examples:
Input: N = 10, A[] = [2, 3, 4, 2, 4, 5, 7, 4, 3, 9], K = 4
Output: 9
Explanation : All Subarrays of size K with all distinct integers are [2, 4, 5, 7], [5, 7, 4, 3], [7, 4, 3, 9] and there XOR are 6, 5, 9 respectively. So Maximum XOR of subarray is 9.
Input: N = 5, A[] = [2, 3, 2, 3, 2], K = 3
Output: -1
Explanation: Since there of no subarray of size K with all integers distinct
Naive Solution: The basic way to solve the problem is as follows:
A Simple Solution is to generate all subarrays of size K, check if all the integers in that subarray are distinct, compute their XORs, and finally return the maximum of all XORs, if no subarray of size K with all distinct is found return -1.
Below is the implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
// Function performing Calculation
int maximumXorsubarray(int N, vector<int>& A, int K)
{
// Variable for storing maximum XOR
// of the subarray of size K
int mx = -1;
// Generating all subarray of size K
for (int i = 0; i < N - K + 1; i++) {
// Set is used for storing the
// element of subarray and
// checking distinct condition
unordered_set<int> st;
for (int j = i; j < i + K; j++) {
st.insert(A[j]);
}
// If subarray is of size K with
// all distinct
if (st.size() == K) {
// Calculating xor of the
// subarray of size K
int xorr = 0;
for (auto it : st) {
xorr = (xorr ^ it);
}
// update mx
mx = max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
// Driver function
int main()
{
// Size of given subarray
int N = 10;
// Given array A
vector<int> A = { 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 };
// Size of subarray needed
int K = 4;
// Function Call
cout << "Maximum XOR of subarray of size K with all "
"distincts are : "
<< maximumXorsubarray(N, A, K);
return 0;
}
// Java code for the above approach:
import java.util.*;
class GFG {
// Function performing Calculation
static int maximumXorsubarray(int N,int A[], int K)
{
// Variable for storing maximum XOR
// of the subarray of size K
int mx = -1;
// Generating all subarray of size K
for (int i = 0; i < N - K + 1; i++) {
// Set is used for storing the
// element of subarray and
// checking distinct condition
Set<Integer> st = new HashSet<>();
for (int j = i; j < i + K; j++) {
st.add(A[j]);
}
// If subarray is of size K with
// all distinct
if (st.size() == K) {
// Calculating xor of the
// subarray of size K
int xorr = 0;
for (Integer it : st) {
xorr = (xorr ^ it);
}
// update mx
mx = Math.max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
// Driver function
public static void main (String[] args) {
// Size of given subarray
int N = 10;
// Given array A
int A[] = { 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 };
// Size of subarray needed
int K = 4;
// Function Call
System.out.println( "Maximum XOR of subarray of size K with all distincts are : " + maximumXorsubarray(N, A, K));
}
}
# Python3 code for the above approach
from typing import List
# Function performing Calculation
def maximumXorsubarray(N: int, A: List[int], K: int) -> int:
# Variable for storing maximum XOR
# of the subarray of size K
mx = -1
# Generating all subarray of size K
for i in range(N - K + 1):
# Set is used for storing the
# element of subarray and
# checking distinct condition
st = set()
for j in range(i, i + K):
st.add(A[j])
# If subarray is of size K with
# all distinct
if len(st) == K:
# Calculating xor of the
# subarray of size K
xorr = 0
for it in st:
xorr = (xorr ^ it)
# update mx
mx = max(mx, xorr)
# Returning the Maximum XOR of subarray
# of size K with all distinct
return mx
# Driver function
if __name__ == "__main__":
# Size of given subarray
N = 10
# Given array A
A = [2, 3, 4, 2, 4, 5, 7, 4, 3, 9]
# Size of subarray needed
K = 4
# Function Call
print("Maximum XOR of subarray of size K with all distincts are : ", maximumXorsubarray(N, A, K))
# This code is contributed by lokeshpotta20.
// C# code for the above approach
using System;
using System.Collections.Generic;
public class GFG {
// Function performing Calculation
static int MaximumXorSubarray(int N, int[] A, int K)
{
// Variable for storing maximum XOR of the subarray
// of size K
int mx = -1;
// Generating all subarray of size K
for (int i = 0; i < N - K + 1; i++)
{
// Set is used for storing the element of
// subarray and checking distinct condition
HashSet<int> st = new HashSet<int>();
for (int j = i; j < i + K; j++) {
st.Add(A[j]);
}
// If subarray is of size K with all distinct
if (st.Count == K)
{
// Calculating xor of the subarray of size K
int xorr = 0;
foreach(int it in st)
{
xorr = (xorr ^ it);
}
// update mx
mx = Math.Max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray of size K
// with all distinct
return mx;
}
static public void Main()
{
// Code
// Size of given subarray
int N = 10;
// Given array A
int[] A = { 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 };
// Size of subarray needed
int K = 4;
// Function Call
Console.WriteLine(
"Maximum XOR of subarray of size K with all distincts are : "
+ MaximumXorSubarray(N, A, K));
}
}
// This code is contributed by sankar.
// Function performing Calculation
function maximumXorsubarray(N, A, K) {
// Variable for storing maximum XOR
// of the subarray of size K
let mx = -1;
// Generating all subarray of size K
for (let i = 0; i <= N - K; i++) {
// Set is used for storing the
// element of subarray and
// checking distinct condition
let st = new Set();
for (let j = i; j < i + K; j++) {
st.add(A[j]);
}
// If subarray is of size K with
// all distinct
if (st.size === K) {
// Calculating xor of the
// subarray of size K
let xorr = 0;
for (let it of st) {
xorr = (xorr ^ it);
}
// update mx
mx = Math.max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
// Driver function
const N = 10;
const A = [2, 3, 4, 2, 4, 5, 7, 4, 3, 9];
const K = 4;
// Function Call
console.log("Maximum XOR of subarray of size K with all distincts are : ", maximumXorsubarray(N, A, K));
OutputMaximum XOR of subarray of size K with all distincts are : 9
Time Complexity: O(N*K)
Auxiliary Space: O(K)
Efficient Approach: To solve the problem follow the below idea:
In this approach, we will use the XOR property that ( X XOR X ) is 0 and we have to take a map that will store the frequencies of the integers and use the 2-pointer approach, whenever we get the element whose frequency exceeds 1 or whenever the size of the map will exceed K then we will shrink window otherwise we will expand the window.
Below is the implementation of the above approach:
C++
// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
// Function performing Calculation
int maximumXorsubarray(int N, vector<int>& A, int K)
{
// Variable for storing maximum XOR
// of the subarray of size K
int mx = -1;
// Declaring map which is used for
// storing frequencies
map<int, int> mp;
// Temporary variable
int xorr = 0;
// Using 2-pointers i, j
int i = 0;
for (int j = 0; j < N; j++) {
// Expanding the window
mp[A[j]]++;
xorr = (xorr xor A[j]);
// Shrinking the window
while (mp[A[j]] > 1 || mp.size() > K) {
mp[A[i]]--;
xorr = (xorr xor A[i]);
if (mp[A[i]] == 0)
mp.erase(A[i]);
i++;
}
// If size of window is equal to K
// then updating the mx variable
if ((j - i + 1) == K) {
mx = max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
// Driver function
int main()
{
// Size of given subarray
int N = 10;
// Given array A
vector<int> A = { 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 };
// Size of subarray needed
int K = 4;
// Function Call
cout << "Maximum XOR of subarray of size K with all "
"distinct are : "
<< maximumXorsubarray(N, A, K);
return 0;
}
// Java code for the above approach:
import java.util.*;
class GFG {
// Function performing Calculation
static int maximumXorsubarray(int N, int[] A, int K) {
// Variable for storing maximum XOR
// of the subarray of size K
int mx = -1;
// Declaring map which is used for
// storing frequencies
Map<Integer, Integer> mp = new HashMap<>();
// Temporary variable
int xorr = 0;
// Using 2-pointers i, j
int i = 0;
for (int j = 0; j < N; j++) {
// Expanding the window
mp.put(A[j], mp.getOrDefault(A[j], 0) + 1);
xorr = xorr ^ A[j];
// Shrinking the window
while (mp.get(A[j]) > 1 || mp.size() > K) {
int count = mp.get(A[i]);
count--;
if (count == 0) {
mp.remove(A[i]);
} else {
mp.put(A[i], count);
}
xorr = xorr ^ A[i];
i++;
}
// If size of window is equal to K
// then updating the mx variable
if ((j - i + 1) == K) {
mx = Math.max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
// Driver function
public static void main(String[] args) {
// Size of given subarray
int N = 10;
// Given array A
int[] A = {2, 3, 4, 2, 4, 5, 7, 4, 3, 9};
// Size of subarray needed
int K = 4;
// Function Call
System.out.println("Maximum XOR of subarray of size K with all " +
"distinct are: " + maximumXorsubarray(N, A, K));
}
}
// This Code is Contributed by Prasad Kandekar(prasad264)
// C# code for the above approach:
using System;
using System.Collections.Generic;
public class GFG {
// Function performing Calculation
static int maximumXorsubarray(int N, int[] A, int K)
{
// Variable for storing maximum XOR
// of the subarray of size K
int mx = -1;
// Declaring map which is used for
// storing frequencies
Dictionary<int, int> mp
= new Dictionary<int, int>();
// Temporary variable
int xorr = 0;
// Using 2-pointers i, j
int i = 0;
for (int j = 0; j < N; j++) {
// Expanding the window
if (mp.ContainsKey(A[j])) {
mp[A[j]]++;
}
else {
mp[A[j]] = 1;
}
xorr = xorr ^ A[j];
// Shrinking the window
while (mp[A[j]] > 1 || mp.Count > K) {
int count = mp[A[i]];
count--;
if (count == 0) {
mp.Remove(A[i]);
}
else {
mp[A[i]] = count;
}
xorr = xorr ^ A[i];
i++;
}
// If size of window is equal to K
// then updating the mx variable
if ((j - i + 1) == K) {
mx = Math.Max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct
return mx;
}
static public void Main()
{
// Code
// Size of given subarray
int N = 10;
// Given array A
int[] A = { 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 };
// Size of subarray needed
int K = 4;
// Function Call
Console.WriteLine(
"Maximum XOR of subarray of size K with all "
+ "distinct are: "
+ maximumXorsubarray(N, A, K));
}
}
// This code is contributed by karthik.
def maximumXorsubarray(N, A, K):
# Variable for storing maximum XOR
# of the subarray of size K
mx = -1
# Declaring map which is used for
# storing frequencies
mp = {}
# Temporary variable
xorr = 0
# Using 2-pointers i, j
i = 0
for j in range(N):
# Expanding the window
if A[j] not in mp:
mp[A[j]] = 0
mp[A[j]] += 1
xorr ^= A[j]
# Shrinking the window
while mp[A[j]] > 1 or len(mp) > K:
mp[A[i]] -= 1
xorr ^= A[i]
if mp[A[i]] == 0:
mp.pop(A[i])
i += 1
# If size of window is equal to K
# then updating the mx variable
if j - i + 1 == K:
mx = max(mx, xorr)
# Returning the Maximum XOR of subarray
# of size K with all distinct
return mx
# Size of given subarray
N = 10
# Given array A
A = [2, 3, 4, 2, 4, 5, 7, 4, 3, 9]
# Size of subarray needed
K = 4
# Function Call
print("Maximum XOR of subarray of size K with all "
"distinct are: ", maximumXorsubarray(N, A, K))
// JavaScript code for the above approach:
function maximumXorsubarray(N, A, K)
{
// Variable for storing maximum XOR
// of the subarray of size K
let mx = -1;
// Declaring map which is used for
// storing frequencies
let mp = new Map();
// Temporary variable
let xorr = 0;
// Using 2-pointers i, j
let i = 0;
for (let j = 0; j < N; j++) {
// Expanding the window
mp.set(A[j], (mp.get(A[j]) || 0) + 1);
xorr ^= A[j];
// Shrinking the window
while (mp.get(A[j]) > 1 || mp.size > K) {
mp.set(A[i], mp.get(A[i]) - 1);
xorr ^= A[i];
if (mp.get(A[i]) == 0)
mp.delete(A[i]);
i++;
}
// If size of window is equal to K
// then updating the mx variable
if (j - i + 1 == K) {
mx = Math.max(mx, xorr);
}
}
// Returning the Maximum XOR of subarray
// of size K with all distinct elements
return mx;
}
// Driver function
const N = 10;
// Given array A
const A = [ 2, 3, 4, 2, 4, 5, 7, 4, 3, 9 ];
// Size of subarray needed
const K = 4;
// Function Call
console.log("Maximum XOR of subarray of size K with all distinct elements is: "
+ maximumXorsubarray(N, A, K));
OutputMaximum XOR of subarray of size K with all distict are : 9
Time Complexity: O(N*Log(K))
Auxiliary Space: O(K)
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