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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: 

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Input: arr[] = { 1, 0, 2, 3} 
Output:
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: 

  1. Insert all the elements into the set.
  2. Iterate for all possible values of K in the range [0, 1024].
  3. For every element in the set, find its Bitwise XOR with K.
  4. 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.
  5. 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
1

 

Time Complexity: O(K*N*log2N) 
Auxiliary Space: O(1)
 




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