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

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, 2pos will be added to the final value.

    For each recursive step:

    answer = min(answer_on, answer_off) + 2pos

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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    

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C#

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// 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

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Output:

5

Time Complexity: O(N * log(max(Ai))

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Improved By : chitranayal, princi singh