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Find a number such that maximum in array is minimum possible after XOR

  • Last Updated : 29 Apr, 2021

Given an array of non-negative integers. Choose an integer P and take XOR of P with all elements of the array. The task is to choose P such that the maximum value of the array is minimum possible after performing XOR of all elements of the array with P.

Examples:

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Input: arr = {3, 2, 1} 
Output:
Explanation: 
We can choose P = 3 such that after taking XOR the maximum element is minimum possible. 
After taking exclusive-OR arr = {0, 1, 2} 
2 is minimum possible value.
Input: arr = {5, 1} 
Output: 4



Approach:

  • We can solve this problem recursively starting from the most significant bit.
  • Split the elements into two groups, one with the elements which have the current bit on (1) and one with the elements which have the current bit off (0).
  • If either group is empty, we can assign the current bit of P accordingly so that we have the current bit off in our answer.
  • Otherwise, If both groups aren’t empty, so whatever value we assign to the current bit of P, we will have this bit on(1) in our answer.
  • Now, To decide which value to assign to the current bit of P, we will recursively call for each of the groups for the next bit and return the minimum of both.

Below is the implementation of the above approach:

C++




// C++ program that find the minimum
// possible maximum
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function that find the
//  minimum value after exclusive-OR
int RecursiveFunction(vector<int> ref,
                      int bit)
{
    // Condition if ref size is zero or
    // bit is negative then return 0
    if (ref.size() == 0 || bit < 0)
        return 0;
 
    vector<int> curr_on, curr_off;
 
    for (int i = 0; i < ref.size(); i++)
    {
        // Condition if current bit is
        // off then push current value
        // in curr_off vector
        if (((ref[i] >> bit) & 1) == 0)
            curr_off.push_back(ref[i]);
         
        // Condition if current bit is on
        // then push current value in
        // curr_on vector
        else
            curr_on.push_back(ref[i]);
    }
 
    // Condition if curr_off is empty
    // then call recursive function
    // on curr_on vector
    if (curr_off.size() == 0)
        return RecursiveFunction(curr_on,
                                 bit - 1);
 
    // Condition if curr_on is empty
    // then call recursive function
    // on curr_off vector
    if (curr_on.size() == 0)
        return RecursiveFunction(curr_off,
                                 bit - 1);
 
    // Return the minimum of curr_off and 
    // curr_on and add power of 2 of
    // current bit
    return min(RecursiveFunction(curr_off,
                                 bit - 1),
               RecursiveFunction(curr_on,
                                 bit - 1))
           + (1 << bit);
}
 
// Function that print the minimum
// value after exclusive-OR
void PrintMinimum(int a[], int n)
{
    vector<int> v;
 
    // Pushing values in vector
    for (int i = 0; i < n; i++)
        v.push_back(a[i]);
 
    // Printing answer
    cout << RecursiveFunction(v, 30)
         << "\n";
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 1 };
 
    int size = sizeof(arr) / sizeof(arr[0]);
 
    PrintMinimum(arr, size);
 
    return 0;
}

Java




// Java program that find the minimum
// possible maximum
import java.util.*;
 
class GFG{
 
// Recursive function that find the
// minimum value after exclusive-OR
static int RecursiveFunction(ArrayList<Integer> ref,
                             int bit)
{
     
    // Condition if ref size is zero or
    // bit is negative then return 0
    if (ref.size() == 0 || bit < 0)
        return 0;
 
    ArrayList<Integer> curr_on = new ArrayList<>();
    ArrayList<Integer> curr_off = new ArrayList<>();
 
    for(int i = 0; i < ref.size(); i++)
    {
         
        // Condition if current bit is
        // off then push current value
        // in curr_off vector
        if (((ref.get(i) >> bit) & 1) == 0)
            curr_off.add(ref.get(i));
 
        // Condition if current bit is on
        // then push current value in
        // curr_on vector
        else
            curr_on.add(ref.get(i));
    }
 
    // Condition if curr_off is empty
    // then call recursive function
    // on curr_on vector
    if (curr_off.size() == 0)
        return RecursiveFunction(curr_on, bit - 1);
 
    // Condition if curr_on is empty
    // then call recursive function
    // on curr_off vector
    if (curr_on.size() == 0)
        return RecursiveFunction(curr_off, bit - 1);
 
    // Return the minimum of curr_off and
    // curr_on and add power of 2 of
    // current bit
    return Math.min(RecursiveFunction(curr_off,
                                      bit - 1),
                    RecursiveFunction(curr_on,
                                      bit - 1)) +
                                     (1 << bit);
}
 
// Function that print the minimum
// value after exclusive-OR
static void PrintMinimum(int a[], int n)
{
    ArrayList<Integer> v = new ArrayList<>();
 
    // Pushing values in vector
    for(int i = 0; i < n; i++)
        v.add(a[i]);
 
    // Printing answer
    System.out.println(RecursiveFunction(v, 30));
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 1 };
    int size = arr.length;
 
    PrintMinimum(arr, size);
}
}
 
// This code is contributed by jrishabh99

Python3




# Python3 program that find
# the minimum  possible maximum
 
# Recursive function that find the
#  minimum value after exclusive-OR
def RecursiveFunction(ref, bit):
 
    # Condition if ref size is zero or
    # bit is negative then return 0
    if(len(ref) == 0 or bit < 0):
        return 0;
    curr_on = []
    curr_off = []
 
    for i in range(len(ref)):
       
        # Condition if current bit is 
        # off then push current value 
        # in curr_off vector
        if(((ref[i] >> bit) & 1) == 0):
            curr_off.append(ref[i])
 
        # Condition if current bit is on 
        # then push current value in 
        # curr_on vector
        else:
            curr_on.append(ref[i])
 
    # Condition if curr_off is empty
    # then call recursive function
    # on curr_on vector
    if(len(curr_off) == 0):
        return RecursiveFunction(curr_on,
                                 bit - 1)
 
    # Condition if curr_on is empty
    # then call recursive function
    # on curr_off vector
    if(len(curr_on) == 0):
        return RecursiveFunction(curr_off,
                                 bit - 1)
 
    # Return the minimum of curr_off and  
    # curr_on and add power of 2 of
    # current bit
    return(min(RecursiveFunction(curr_off,
                                 bit - 1),
               RecursiveFunction(curr_on,
                                 bit - 1)) + (1 << bit))
 
# Function that print the minimum
# value after exclusive-OR
def PrintMinimum(a, n):
    v = []
 
    # Pushing values in vector
    for i in range(n):
        v.append(a[i])
 
    # Printing answer
    print(RecursiveFunction(v, 30))
 
# Driver Code
arr = [3, 2, 1]
size = len(arr)
PrintMinimum(arr, size)
 
#This code is contributed by avanitrachhadiya2155

C#




// C# program that find the minimum
// possible maximum
using System;
using System.Collections.Generic;
 
class GFG{
 
// Recursive function that find the
// minimum value after exclusive-OR
static int RecursiveFunction(List<int> re,
                             int bit)
{
     
    // Condition if ref size is zero or
    // bit is negative then return 0
    if (re.Count == 0 || bit < 0)
        return 0;
 
    List<int> curr_on = new List<int>();
    List<int> curr_off = new List<int>();
 
    for(int i = 0; i < re.Count; i++)
    {
         
        // Condition if current bit is
        // off then push current value
        // in curr_off vector
        if (((re[i] >> bit) & 1) == 0)
            curr_off.Add(re[i]);
 
        // Condition if current bit is on
        // then push current value in
        // curr_on vector
        else
            curr_on.Add(re[i]);
    }
 
    // Condition if curr_off is empty
    // then call recursive function
    // on curr_on vector
    if (curr_off.Count == 0)
        return RecursiveFunction(curr_on,
                                 bit - 1);
 
    // Condition if curr_on is empty
    // then call recursive function
    // on curr_off vector
    if (curr_on.Count == 0)
        return RecursiveFunction(curr_off,
                                 bit - 1);
 
    // Return the minimum of curr_off and
    // curr_on and add power of 2 of
    // current bit
    return Math.Min(RecursiveFunction(curr_off,
                                      bit - 1),
                    RecursiveFunction(curr_on,
                                      bit - 1)) +
                                     (1 << bit);
}
 
// Function that print the minimum
// value after exclusive-OR
static void PrintMinimum(int []a, int n)
{
    List<int> v = new List<int>();
 
    // Pushing values in vector
    for(int i = 0; i < n; i++)
        v.Add(a[i]);
 
    // Printing answer
    Console.WriteLine(RecursiveFunction(v, 30));
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 1 };
    int size = arr.Length;
 
    PrintMinimum(arr, size);
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// Javascript program that find the minimum
// possible maximum
 
// Recursive function that find the
//  minimum value after exclusive-OR
function RecursiveFunction(ref, bit)
{
    // Condition if ref size is zero or
    // bit is negative then return 0
    if (ref.length == 0 || bit < 0)
        return 0;
 
    let curr_on = [], curr_off = [];
 
    for (let i = 0; i < ref.length; i++)
    {
        // Condition if current bit is
        // off then push current value
        // in curr_off vector
        if (((ref[i] >> bit) & 1) == 0)
            curr_off.push(ref[i]);
         
        // Condition if current bit is on
        // then push current value in
        // curr_on vector
        else
            curr_on.push(ref[i]);
    }
 
    // Condition if curr_off is empty
    // then call recursive function
    // on curr_on vector
    if (curr_off.length == 0)
        return RecursiveFunction(curr_on,
                                 bit - 1);
 
    // Condition if curr_on is empty
    // then call recursive function
    // on curr_off vector
    if (curr_on.length == 0)
        return RecursiveFunction(curr_off,
                                 bit - 1);
 
    // Return the minimum of curr_off and 
    // curr_on and add power of 2 of
    // current bit
    return Math.min(RecursiveFunction(curr_off,
                                 bit - 1),
               RecursiveFunction(curr_on,
                                 bit - 1))
           + (1 << bit);
}
 
// Function that print the minimum
// value after exclusive-OR
function PrintMinimum(a, n)
{
    let v = [];
 
    // Pushing values in vector
    for (let i = 0; i < n; i++)
        v.push(a[i]);
 
    // Printing answer
    document.write(RecursiveFunction(v, 30)
         + "<br>");
}
 
// Driver Code
    let arr = [ 3, 2, 1 ];
 
    let size = arr.length;
 
    PrintMinimum(arr, size);
 
</script>
Output: 
2



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