Minimize K whose XOR with given array elements leaves array unchanged
Last Updated :
12 May, 2021
Given an array of N elements, the task is to find the minimum value of K such that Bitwise XOR of K with all array elements produces the same set of elements. If it is impossible to find any value of K then print "-1".
Examples:
Input: arr[] = { 1, 0, 2, 3}
Output: 1
Explanation:
For K = 1,
1 xor 1 = 0
1 xor 0 = 1
1 xor 2 = 3
1 xor 3 = 2
Thus, K = 1 is the least possible positive value which leaves the array unaltered.
Input: arr[] = { 7, 1, 2, 3, 8}
Output: -1
Naive Approach: The naive approach is to iterate for all the possible values of K in the range [1, 1024] and check if Bitwise XOR of K with all the elements in the array gives the same array elements or not. If for any minimum value of K the Bitwise XOR produces the same array then print that value of K else print "-1".
Time Complexity: O(K*N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by using additional space. Below are the steps:
- Insert all the elements into the set.
- Iterate for all possible values of K in the range [0, 1024].
- For every element in the set, find its Bitwise XOR with K.
- The first value of K for which all the elements generated after Bitwise XOR with the element inset is the same as that of the given set, then print the value of K.
- If no such K is obtained, print "-1".
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum
// value of K in given range
int min_value(int arr[], int N)
{
int x, X, K;
// Declare a set
set<int> S;
for (int i = 0; i < N; i++) {
S.insert(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
for (auto it = S.begin(); it != S.end(); it++)
// Check if the XOR
// calculated is present
// in the Set
{
X = ((i | *it) - (i & *it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.find(X) != S.end()) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.size()) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 0, 3, 3, 0, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << min_value(arr, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG {
// Function to find the minimum
// value of K in given range
static int min_value(int arr[], int N)
{
int x, X, K;
// Declare a set
HashSet<Integer> S = new HashSet<Integer>();
for (int i = 0; i < N; i++) {
S.add(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
for (int it : S) {
// Check if the XOR
// calculated is present
// in the Set
X = ((i | it) - (i & it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.contains(X)) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.size()) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 0, 3, 3, 0, 2 };
int N = arr.length;
// Function Call
System.out.print(min_value(arr, N));
}
}
// This code is contributed by Rajput-Ji
# Python3 program for the above approach
# Function to find the minimum
# value of K in given range
def min_value(arr, N):
x, X, K = 0, 0, 0
# Declare a set
S = set()
for i in range(N):
S.add(arr[i])
# Initialize count variable
count = 0
# Iterate in range [1, 1024]
for i in range(1, 1024):
# counter set as 0
count = 0
# Iterating through the Set
for it in S:
# Check if the XOR
# calculated is present
# in the Set
X = ((i | it) - (i & it))
# If the value of Bitwise XOR
# inside the given set then
# increment count
if X in S:
count += 1
# Check if the value of count is
# equal to the size of set
if (count == len(S)):
K = i
# Return minimum value of K
return K
# If no such K is found
return -1
# Driver Code
# Given array
arr = [1, 0, 3, 3, 0, 2]
N = len(arr)
# Function Call
print(min_value(arr, N))
# This code is contributed by shubhamsingh10
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find the minimum
// value of K in given range
static int min_value(int[] arr, int N)
{
// int x;
int X, K;
// Declare a set
HashSet<int> S = new HashSet<int>();
for (int i = 0; i < N; i++) {
S.Add(arr[i]);
}
// Initialize count variable
int count = 0;
// Iterate in range [1, 1024]
for (int i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
foreach(int it in S)
{
// Check if the XOR
// calculated is present
// in the Set
X = ((i | it) - (i & it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
if (S.Contains(X)) {
count++;
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.Count) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver code
static void Main()
{
// Given array
int[] arr = { 1, 0, 3, 3, 0, 2 };
int N = arr.Length;
// Function call
Console.Write(min_value(arr, N));
}
}
// This code is contributed by divyeshrabadiya07
<script>
// Javascript implementation for the above approach
// Function to find the minimum
// value of K in given range
function min_value(arr, N)
{
let x, X, K;
// Declare a set
let S = [];
for (let i = 0; i < N; i++) {
S.push(arr[i]);
}
// Initialize count variable
let count = 0;
// Iterate in range [1, 1024]
for (let i = 1; i <= 1024; i++) {
// counter set as 0
count = 0;
// Iterating through the Set
for (let it in S) {
// Check if the XOR
// calculated is present
// in the Set
X = ((i | it) - (i & it));
// If the value of Bitwise XOR
// inside the given set then
// increment count
for (let j in S) {
if (S[j]==X) {
count++;
}
}
}
// Check if the value of count is
// equal to the size of set
if (count == S.length) {
K = i;
// Return minimum value of K
return K;
}
}
// If no such K is found
return -1;
}
// Driver Code
// Given array
let arr = [ 1, 0, 3, 3, 0, 2 ];
let N = arr.length;
// Function Call
document.write(min_value(arr, N));
</script>
Time Complexity: O(K*N*log2N)
Auxiliary Space: O(1)
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