Maximum Xor


Problem Statement :


You are given an array arr of n elements. A list of integers, queries  is given as an input, find the maximum value of queries[ j ] each arr[ i ] for all  0 <=  i  < n, where  represents xor of two elements.

Note that there are multiple test cases in one input file.


Function Description

Complete the maxXor function in the editor below. It must return an array of integers, each representing the maximum xor value for each element queries[ j ] against all elements of arr.

maxXor has the following parameter(s):

arr: an array of integers
queries: an array of integers to query



Input Format


The first line contains an integer n, the size of the array arr.

The second line contains n space-separated integers, arr[ i ] from 0  <= i  < n.

The third line contain m, the size of the array queries.

Each of the next m lines contains an integer queries[ j ] where 0  <= j  < m.



Constraints


1  <=   n, m  <=  10^5
0   <=  arr[ i ] , queries[ j ]  <= 10^9


Output Format

The output should contain m lines with each line representing output for the corresponding input of the testcase.


Solution :



title-img



                        Solution in C++ :

In   C++  :








#include <bits/stdc++.h>
using namespace std;

int p;
struct node
{
    int val;
    node *child[2];
};

void insert(node *trie, int x, int ind)
{   
    if(ind < 0) {
        return;
    }
    int k=(x>>ind)&1;

    if(!trie->child[k]) {
        trie->child[k]=new node;
    }
    insert(trie->child[k], x, ind-1);
}

void find(node *trie, int x, int ind)
{   
    if(ind<0) {
        return;
    }

    int k=(x>>ind)&1;
    k^=1;

    if(!trie->child[k]) {
        k^=1;
    }
    p=p<<1|k;
    find(trie->child[k], x, ind-1);
}

int main()
{
    int n,i,x;
    cin>>n;

    int a[n];
    for(i=0;i<n;++i) {
        cin>>a[i];
    }

    node *trie = new node;
    for(i=0;i<n;++i) {
        // max 32 bits
        insert(trie,a[i],31);
    }

    int t;
    cin>>t;
    while(t--) {
        cin>>x;
        p=0;
        find(trie,x,31);
        cout<<(p^x)<<endl;
    }

    return 0;
}
                    

                        Solution in Java :

In   Java   :









static final int INT_SIZE = 32; 

// A Trie Node 
static class TrieNode 
{ 
    int value;  // Only used in leaf nodes 
    TrieNode[] arr =  new TrieNode[2]; 
    public TrieNode() { 
        value = 0; 
        arr[0] = null; 
        arr[1] = null; 
    } 

    public String toString(){
        return "{ value: "+this.value+" "+ arr.toString()+"}";
    }
} 
static TrieNode root;

static void insert(int pre_xor) 
{ 
    TrieNode temp = root; 

    for (int i=INT_SIZE-1; i>=0; i--) 
    { 
        int val = (pre_xor & (1<<i)) >=1 ? 1 : 0; 

        if (temp.arr[val] == null) 
            temp.arr[val] = new TrieNode(); 

        temp = temp.arr[val]; 
    } 
    temp.value = pre_xor; 
} 

static int query(int pre_xor) 
{ 
    TrieNode temp = root; 
    for (int i=INT_SIZE-1; i>=0; i--) 
    { 
        int val = (pre_xor & (1<<i)) >= 1 ? 1 : 0; 

        if (temp.arr[val] != null) 
            temp = temp.arr[val]; 

        else if (temp.arr[1-val] != null) 
            temp = temp.arr[1-val];
    } 
    return (~pre_xor)^(temp.value); 
}
// Complete the maxXor function below.
static int[] maxXor(int[] arr, int[] queries) {
    int[] result = new int[queries.length];

    root = new TrieNode(); 
    insert(0);
    int pre_xor =0; 
    int max = Integer.MIN_VALUE;
    for (int i=0; i<arr.length; i++) 
    {  
        insert(arr[i]);
    }  

    for(int j=0; j<queries.length;j++){
        pre_xor = ~queries[j];  
        //max= Math.max(max, query(pre_xor));
        result[j]=query(pre_xor);
    }
    return result;
}
                    

                        Solution in Python : 
                            
In   Python3  :







def maxXor(arr, queries):
    ans = []
    trie = {}
    k = len(bin(max(arr+queries))) - 2 
    for number in ['{:b}'.format(x).zfill(k) for x in arr]:
        node = trie
        for char in number:
            node = node.setdefault(char, {})
    for n in queries:
        node = trie
        s = ''
        for char in'{:b}'.format(n).zfill(k) :
            tmp = str(int(char) ^ 1) 
            tmp = tmp if tmp in node else char 
            s += tmp 
            node = node[tmp]
        ans.append(int(s, 2) ^ n) 
    return ans
                    

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