Choose an integer K such that maximum of the xor values of K with all Array elements is minimized

Last Updated : 15 Feb, 2023

Given an array A consisting of N non-negative integers, the task is to choose an integer K such that the maximum of the xor values of K with all array elements is minimized. In other words find the minimum possible value of Z, where Z = max(A[i] xor K), 0 <= i <= n-1, for some value of K.
Examples:

Input: A = [1, 2, 3]
Output: 2
Explanation:
On choosing K = 3, max(A[i] xor 3) = 2, and this is the minimum possible value.
Input: A = [3, 2, 5, 6]
Output: 5

Approach: To solve the problem mentioned above we will use recursion. We will start from the most significant bit in the recursive function.

In the recursive step, split the element into two sections – one having the current bit on and the other with current bit off. If any of the sections doesn’t have a single element, then this particular bit for K can be chosen such that the final xor value has 0 at this bit position (since our aim is to minimise this value) and then proceed to the next bit in the next recursive step.
If both the sections have some elements, then explore both the possibilities by placing 0 and 1 at this bit position and calculating the answer using the corresponding section in next recursive call.
Let answer_on be the value if 1 is placed and answer_off be the value if 0 is placed at this position (pos). Since both sections are non empty whichever bit we choose for K, 2^pos will be added to the final value.
For each recursive step:

answer = min(answer_on, answer_off) + 2^pos

Below is the implementation of the above approach:

C++

// C++ implementation to find Minimum
// possible value of the maximum xor
// in an array by choosing some integer
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate Minimum possible
// value of the Maximum XOR in an array
int calculate(vector<int>& section, int pos)
{
    // base case
    if (pos < 0)
        return 0;
 
    // Divide elements into two sections
    vector<int> on_section, off_section;
 
    // Traverse all elements of current
    // section and divide in two groups
    for (auto el : section) {
        if (((el >> pos) & 1) == 0)
            off_section.push_back(el);
 
        else
            on_section.push_back(el);
    }
 
    // Check if one of the sections is empty
    if (off_section.size() == 0)
        return calculate(on_section, pos - 1);
 
    if (on_section.size() == 0)
        return calculate(off_section, pos - 1);
 
    // explore both the possibilities using recursion
    return min(calculate(off_section, pos - 1),
               calculate(on_section, pos - 1))
           + (1 << pos);
}
 
// Function to calculate minimum XOR value
int minXorValue(int a[], int n)
{
    vector<int> section;
    for (int i = 0; i < n; i++)
        section.push_back(a[i]);
 
    // Start recursion from the
    // most significant pos position
    return calculate(section, 30);
}
 
// Driver code
int main()
{
    int N = 4;
 
    int A[N] = { 3, 2, 5, 6 };
 
    cout << minXorValue(A, N);
 
    return 0;
}

Java

// Java implementation to find Minimum
// possible value of the maximum xor
// in an array by choosing some integer
import java.util.*;
 
class GFG{
 
// Function to calculate Minimum possible
// value of the Maximum XOR in an array
static int calculate(Vector<Integer> section, int pos)
{
 
    // Base case
    if (pos < 0)
        return 0;
 
    // Divide elements into two sections
    Vector<Integer> on_section = new Vector<Integer>(), 
                   off_section = new Vector<Integer>();
 
    // Traverse all elements of current
    // section and divide in two groups
    for(int el : section) 
    {
       if (((el >> pos) & 1) == 0)
           off_section.add(el);
       else
           on_section.add(el);
    }
 
    // Check if one of the sections is empty
    if (off_section.size() == 0)
        return calculate(on_section, pos - 1);
 
    if (on_section.size() == 0)
        return calculate(off_section, pos - 1);
 
    // Explore both the possibilities using recursion
    return Math.min(calculate(off_section, pos - 1),
                    calculate(on_section, pos - 1)) +
                             (1 << pos);
}
 
// Function to calculate minimum XOR value
static int minXorValue(int a[], int n)
{
    Vector<Integer> section = new Vector<Integer>();
 
    for(int i = 0; i < n; i++)
       section.add(a[i]);
 
    // Start recursion from the
    // most significant pos position
    return calculate(section, 30);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int A[] = { 3, 2, 5, 6 };
 
    System.out.print(minXorValue(A, N));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation to find Minimum
# possible value of the maximum xor
# in an array by choosing some integer
  
# Function to calculate Minimum possible
# value of the Maximum XOR in an array
 
def calculate(section, pos):
 
    # base case
    if (pos < 0):
        return 0
  
    # Divide elements into two sections
    on_section = []
    off_section = []
  
    # Traverse all elements of current
    # section and divide in two groups
    for el in section:
        if (((el >> pos) & 1) == 0):
            off_section.append(el)
  
        else:
            on_section.append(el)
  
    # Check if one of the sections is empty
    if (len(off_section) == 0):
        return calculate(on_section, pos - 1)
  
    if (len(on_section) == 0):
        return calculate(off_section, pos - 1)
  
    # explore both the possibilities using recursion
    return min(calculate(off_section, pos - 1),
               calculate(on_section, pos - 1))+ (1 << pos)
  
# Function to calculate minimum XOR value
def minXorValue(a, n):
    section = []
    for i in range( n):
        section.append(a[i]);
  
    # Start recursion from the
    # most significant pos position
    return calculate(section, 30)
  
# Driver code
if __name__ == "__main__":
    N = 4
  
    A = [ 3, 2, 5, 6 ]
  
    print(minXorValue(A, N))
  
# This code is contributed by chitranayal    

C#

// C# implementation to find minimum
// possible value of the maximum xor
// in an array by choosing some integer
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to calculate minimum possible
// value of the maximum XOR in an array
static int calculate(List<int> section, int pos)
{
     
    // Base case
    if (pos < 0)
        return 0;
 
    // Divide elements into two sections
    List<int> on_section = new List<int>(), 
             off_section = new List<int>();
 
    // Traverse all elements of current
    // section and divide in two groups
    foreach(int el in section) 
    {
        if (((el >> pos) & 1) == 0)
            off_section.Add(el);
        else
            on_section.Add(el);
    }
 
    // Check if one of the sections is empty
    if (off_section.Count == 0)
        return calculate(on_section, pos - 1);
 
    if (on_section.Count == 0)
        return calculate(off_section, pos - 1);
 
    // Explore both the possibilities using recursion
    return Math.Min(calculate(off_section, pos - 1),
                    calculate(on_section, pos - 1)) +
                             (1 << pos);
}
 
// Function to calculate minimum XOR value
static int minXorValue(int []a, int n)
{
    List<int> section = new List<int>();
 
    for(int i = 0; i < n; i++)
       section.Add(a[i]);
 
    // Start recursion from the
    // most significant pos position
    return calculate(section, 30);
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 4;
    int []A = { 3, 2, 5, 6 };
 
    Console.Write(minXorValue(A, N));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
// Javascript implementation to find Minimum
// possible value of the maximum xor
// in an array by choosing some integer
 
// Function to calculate Minimum possible
// value of the Maximum XOR in an array
function calculate(section, pos)
{
    // base case
    if (pos < 0)
        return 0;
 
    // Divide elements into two sections
    let on_section = [], off_section = [];
 
    // Traverse all elements of current
    // section and divide in two groups
    for (let el = 0; el < section.length; el++) {
        if (((section[el] >> pos) & 1) == 0)
            off_section.push(section[el]);
 
        else
            on_section.push(section[el]);
    }
 
    // Check if one of the sections is empty
    if (off_section.length == 0)
        return calculate(on_section, pos - 1);
 
    if (on_section.length == 0)
        return calculate(off_section, pos - 1);
 
    // explore both the possibilities using recursion
    return Math.min(calculate(off_section, pos - 1),
               calculate(on_section, pos - 1))
           + (1 << pos);
}
 
// Function to calculate minimum XOR value
function minXorValue(a, n)
{
    let section = [];
    for (let i = 0; i < n; i++)
        section.push(a[i]);
 
    // Start recursion from the
    // most significant pos position
    return calculate(section, 30);
}
 
// Driver code
    let N = 4;
 
    let A = [ 3, 2, 5, 6 ];
 
    document.write(minXorValue(A, N));
 
</script>

Output:

Time Complexity: O(N * log(max(A_i))
Space complexity: O(NlogN)

Suggest improvement

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Choose an integer K such that maximum of the xor values of K with all Array elements is minimized

C++

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