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.Â
Examples:Â
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Input: A = [1, 2, 3]Â
Output: 2Â
Explanation:Â
On choosing K = 3, max(A[i] xor 3) = 2, and this is the minimum possible value.
Input: A = [3, 2, 5, 6]Â
Output: 5Â
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Approach: To solve the problem mentioned above we will use recursion. We will start from the most significant bit in the recursive function.Â
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- 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.Â
 - 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.Â
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.Â
For each recursive step:Â
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answer = min(answer_on, answer_off) + 2posÂ
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Below is the implementation of the above approach:Â
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C++
// C++ implementation to find Minimum// possible value of the maximum xor// in an array by choosing some integerÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to calculate Minimum possible// value of the Maximum XOR in an arrayint calculate(vector<int>& section, int pos){    // base case    if (pos < 0)        return 0;Â
    // Divide elements into two sections    vector<int> on_section, off_section;Â
    // 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);Â
        else            on_section.push_back(el);    }Â
    // Check if one of the sections is empty    if (off_section.size() == 0)        return calculate(on_section, pos - 1);Â
    if (on_section.size() == 0)        return calculate(off_section, pos - 1);Â
    // explore both the possibilities using recursion    return min(calculate(off_section, pos - 1),               calculate(on_section, pos - 1))           + (1 << pos);}Â
// Function to calculate minimum XOR valueint minXorValue(int a[], int n){Â Â Â Â vector<int> section;Â Â Â Â for (int i = 0; i < n; i++)Â Â Â Â Â Â Â Â section.push_back(a[i]);Â
    // Start recursion from the    // most significant pos position    return calculate(section, 30);}Â
// Driver codeint main(){Â Â Â Â int N = 4;Â
    int A[N] = { 3, 2, 5, 6 };Â
    cout << minXorValue(A, N);Â
    return 0;} |
Java
// Java implementation to find Minimum// possible value of the maximum xor// in an array by choosing some integerimport java.util.*;Â
class GFG{Â
// Function to calculate Minimum possible// value of the Maximum XOR in an arraystatic int calculate(Vector<Integer> section, int pos){Â
    // Base case    if (pos < 0)        return 0;Â
    // Divide elements into two sections    Vector<Integer> on_section = new Vector<Integer>(),                    off_section = new Vector<Integer>();Â
    // 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);    }Â
    // Check if one of the sections is empty    if (off_section.size() == 0)        return calculate(on_section, pos - 1);Â
    if (on_section.size() == 0)        return calculate(off_section, pos - 1);Â
    // Explore both the possibilities using recursion    return Math.min(calculate(off_section, pos - 1),                    calculate(on_section, pos - 1)) +                             (1 << pos);}Â
// Function to calculate minimum XOR valuestatic int minXorValue(int a[], int n){Â Â Â Â Vector<Integer> section = new Vector<Integer>();Â
    for(int i = 0; i < n; i++)       section.add(a[i]);Â
    // Start recursion from the    // most significant pos position    return calculate(section, 30);}Â
// Driver codepublic static void main(String[] args){Â Â Â Â int N = 4;Â Â Â Â int A[] = { 3, 2, 5, 6 };Â
    System.out.print(minXorValue(A, N));}}Â
// 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 arrayÂ
def calculate(section, pos):Â
    # 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 valuedef 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 codeif __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 integerusing System;using System.Collections.Generic;Â
class GFG{Â
// Function to calculate minimum possible// value of the maximum XOR in an arraystatic int calculate(List<int> section, int pos){         // Base case    if (pos < 0)        return 0;Â
    // Divide elements into two sections    List<int> on_section = new List<int>(),              off_section = new List<int>();Â
    // 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);    }Â
    // Check if one of the sections is empty    if (off_section.Count == 0)        return calculate(on_section, pos - 1);Â
    if (on_section.Count == 0)        return calculate(off_section, pos - 1);Â
    // Explore both the possibilities using recursion    return Math.Min(calculate(off_section, pos - 1),                    calculate(on_section, pos - 1)) +                             (1 << pos);}Â
// Function to calculate minimum XOR valuestatic int minXorValue(int []a, int n){Â Â Â Â List<int> section = new List<int>();Â
    for(int i = 0; i < n; i++)       section.Add(a[i]);Â
    // Start recursion from the    // most significant pos position    return calculate(section, 30);}Â
// Driver codepublic static void Main(String[] args){Â Â Â Â int N = 4;Â Â Â Â int []A = { 3, 2, 5, 6 };Â
    Console.Write(minXorValue(A, N));}}Â
// 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 integerÂ
// Function to calculate Minimum possible// value of the Maximum XOR in an arrayfunction calculate(section, pos){    // base case    if (pos < 0)        return 0;Â
    // Divide elements into two sections    let on_section = [], off_section = [];Â
    // 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]);Â
        else            on_section.push(section[el]);    }Â
    // Check if one of the sections is empty    if (off_section.length == 0)        return calculate(on_section, pos - 1);Â
    if (on_section.length == 0)        return calculate(off_section, pos - 1);Â
    // explore both the possibilities using recursion    return Math.min(calculate(off_section, pos - 1),               calculate(on_section, pos - 1))           + (1 << pos);}Â
// Function to calculate minimum XOR valuefunction minXorValue(a, n){Â Â Â Â let section = [];Â Â Â Â for (let i = 0; i < n; i++)Â Â Â Â Â Â Â Â section.push(a[i]);Â
    // Start recursion from the    // most significant pos position    return calculate(section, 30);}Â
// Driver code    let N = 4;Â
    let A = [ 3, 2, 5, 6 ];Â
    document.write(minXorValue(A, N));Â
</script> |
Output:Â
5
Â
Time Complexity: O(N * log(max(Ai))
Space complexity: O(NlogN)
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