LeetCode 3277: Maximum XOR Score Subarray Queries Solution

Master LeetCode problem 3277 (Maximum XOR Score Subarray Queries), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3277. Maximum XOR Score Subarray Queries

Hard

Array Dynamic Programming

Problem Explanation

Explanation:

To solve this problem efficiently, we can use the concept of Trie data structure along with bitwise manipulation. We will iterate through the given queries and build a Trie structure representing the binary representation of the numbers in the array. Then, for each query, we will traverse the Trie to find the maximum XOR score.

Algorithm:

Implement a TrieNode class to represent a node in the Trie.
Implement a Trie class to build the Trie structure from the binary representations of the numbers in the array.
For each query [li, ri]:
- Traverse the Trie starting from the most significant bit.
- At each step, choose the maximum XOR path to maximize the final XOR score.
Repeat this process for all queries and store the results in the answer array.

Time Complexity:

Building the Trie structure: O(n * 32) ~ O(n) where n is the number of elements in the array.
Processing each query: O(32) ~ O(1) since we are dealing with bits.
Total time complexity: O(n + q) where q is the number of queries.

Space Complexity:

Trie structure: O(n * 32) ~ O(n) to store the binary representation of each number.
Additional space for storing the answer array and other variables.
Total space complexity: O(n).

Solution Code

class TrieNode {
    TrieNode[] children;

    public TrieNode() {
        children = new TrieNode[2];
    }
}

class Trie {
    TrieNode root;

    public Trie() {
        root = new TrieNode();
    }

    public void insert(int num) {
        TrieNode curr = root;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (curr.children[bit] == null) {
                curr.children[bit] = new TrieNode();
            }
            curr = curr.children[bit];
        }
    }

    public int findMaxXOR(int num) {
        TrieNode curr = root;
        int maxXOR = 0;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (curr.children[1 - bit] != null) {
                maxXOR |= (1 << i);
                curr = curr.children[1 - bit];
            } else {
                curr = curr.children[bit];
            }
        }
        return maxXOR;
    }
}

public int[] maximizeXor(int[] nums, int[][] queries) {
    Trie trie = new Trie();
    for (int num : nums) {
        trie.insert(num);
    }

    int[] result = new int[queries.length];
    for (int i = 0; i < queries.length; i++) {
        result[i] = trie.findMaxXOR(queries[i][1]) ;
    }
    return result;
}

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