You are given a number of queries Q and each query will be of the following types:
- Query 1 : add(x) This means add x into your data structure.
- Query 2 : maxXOR(y) This means print the maximum possible XOR of y with all the elements already stored in the data structure.
1 <= x, y <= 10^9
1 <= 10^5 <= Q
The data structure begins with only a 0 in it.
Example:
Input: (1 10), (1 13), (2 10), (1 9), (1 5), (2 6) Output: 7 15 Add 10 and 13 to stream. Find maximum XOR with 10, which is 7 Insert 9 and 5 Find maximum XOR with 6 which is 15.
A good way to solve this problem is to use a Trie. A prerequisite for this post is Trie Insert and Search.
Each Trie Node will look following:
struct TrieNode { // We use binary and hence we // need only 2 children TrieNode* Children[2]; bool isLeaf; };
Another thing to handle is that we have to pad the binary equivalent of each input number by a suitable number of zeros to the left before storing them. The maximum possible value of x or y is 10^9 and hence 32 bits will be sufficient.
So how does this work?
Assume we have to insert 3 and 7 into Trie. The Trie starts out with 0 and after these three insertions can be visualized like this:
For simplification, the padding has been done to store each number using 3 bits. Note that in binary:
3 is 011
7 is 111
Now if we have to insert 1 into our Trie, we can note that 1 is 001 and we already have path for 00. So we make a new node for the last set bit and after connecting, we get this:
Now if we have to take XOR with 5 which is 101, we note that for the leftmost bit (position 2), we can choose a 0 starting at the root and thus we go to the left. This is the position 2 and we add 2^2 to the answer.
For position 1, we have a 0 in 5 and we see that we can choose a 1 from our current node. Thus we go right and add 2^1 to the answer.
For position 0, we have a 1 in 5 and we see that we cannot choose a 0 from our current node, thus we go right.
The path taken for 5 is shown above. The answer is thus 2^2 + 2^1 = 6.
// C++ program to find maximum XOR in // a stream of integers #include<bits/stdc++.h> using namespace std; struct TrieNode { TrieNode* children[2]; bool isLeaf; }; // This checks if the ith position in // binary of N is a 1 or a 0 bool check( int N, int i) { return ( bool )(N & (1<<i)); } // Create a new Trie node TrieNode* newNode() { TrieNode* temp = new TrieNode; temp->isLeaf = false ; temp->children[0] = NULL; temp->children[1] = NULL; return temp; } // Inserts x into the Trie void insert(TrieNode* root, int x) { TrieNode* Crawler = root; // padding upto 32 bits for ( int i = 31; i >= 0; i--) { int f = check(x, i); if (! Crawler->children[f]) Crawler->children[f] = newNode(); Crawler = Crawler->children[f]; } Crawler->isLeaf = true ; } // Finds maximum XOR of x with stream of // elements so far. int query2(TrieNode *root, int x) { TrieNode* Crawler = root; // Do XOR from root to a leaf path int ans = 0; for ( int i = 31; i >= 0; i--) { // Find i-th bit in x int f = check(x, i); // Move to the child whose XOR with f // is 1. if ((Crawler->children[f ^ 1])) { ans = ans + (1 << i); // update answer Crawler = Crawler->children[f ^ 1]; } // If child with XOR 1 doesn't exist else Crawler = Crawler->children[f]; } return ans; } // Process x (Add x to the stream) void query1(TrieNode *root, int x) { insert(root, x); } // Driver code int main() { TrieNode* root = newNode(); query1(root, 10); query1(root, 13); cout << query2(root, 10) << endl; query1(root, 9); query1(root, 5); cout << query2(root, 6) << endl; return 0; } |
7 15
The space taken by the Trie is O(n*log(n)). Each query of type 1 takes O(log(n)) time. Each query of type 2 takes O(log(n)) time too. Here n is the largest query number.
Follow up problem: What if we are given three queries instead of two?
1) add(x) This means add x into your data structure (duplicates are allowed).
2) maxXOR(y) This means print the maximum possible XOR of y with all the elements already stored in the data structure.
3) remove(z) This means remove one instance of z from the data structure.
What changes in the Trie solution can achieve this?
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