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Maximum XOR of Two Numbers in an Array in C++
Suppose we have a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. We have to find the maximum result of ai XOR aj, where 0 ≤ i, j < n. So if the input is like [3,10,5,15,2,8], then the output will be 28. The max result will be 5 XOR 25 = 28.
To solve this, we will follow these steps −
Define insertNode(), this will take val and head
curr := head
for i in range 31 to 0
bit := val / (2^i) AND 1
if child[bit] of curr is null, then child[bit] of curr := new node
curr := child[bit] of curr
Define find() method. This will take val and head as input
curr := head, ans := 0
for i in range 31 to 0
bit := val / (2^i) AND 1
if child[bit] of curr is null, then ans := ans OR (2^1)/p>
curr := child[bit] of curr
return ans
From the main method, do the following −
ans := 0
n := size of nums
head := new node
for i in range 0 to n – 1, insertNode(nums[i], head)
for i in range 0 to n – 1, ans := max of ans and find (nums[i], head)
return ans
Example (C++)
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
struct Node{
Node* child[2];
Node(){
child[1] = child[0] = NULL;
}
};
class Solution {
public:
void insertNode(int val, Node* head){
Node* curr = head;
for(int i = 31; i>= 0; i--){
int bit = (val >> i) & 1;
if(!curr->child[bit]){
curr->child[bit] = new Node();
}
curr = curr->child[bit];
}
}
int find(int val, Node* head){
Node* curr = head;
int ans = 0;
for(int i = 31; i>= 0; i--){
int bit = (val >> i) & 1;
if(curr->child[!bit]){
ans |= (1 << i);
curr = curr->child[!bit];
} else {
curr = curr->child[bit];
}
}
return ans;
}
int findMaximumXOR(vector<int>& nums) {
int ans = 0;
int n = nums.size();
Node* head = new Node();
for(int i = 0; i < n; i++){
insertNode(nums[i], head);
}
for(int i = 0; i < n; i++){
ans = max(ans, find(nums[i], head));
}
return ans;
}
};
main(){
vector<int> v = {3,10,5,25,2,8};
Solution ob;
cout << (ob.findMaximumXOR(v));
}Input
[3,10,5,25,2,8]
Output
28
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