LeetCode #3277 — HARD

Maximum XOR Score Subarray Queries

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array nums of n integers, and a 2D integer array queries of size q, where queries[i] = [l_i, r_i].

For each query, you must find the maximum XOR score of any subarray of nums[l_i..r_i].

The XOR score of an array a is found by repeatedly applying the following operations on a so that only one element remains, that is the score:

Simultaneously replace a[i] with a[i] XOR a[i + 1] for all indices i except the last one.
Remove the last element of a.

Return an array answer of size q where answer[i] is the answer to query i.

Example 1:

Input: nums = [2,8,4,32,16,1], queries = [[0,2],[1,4],[0,5]]

Output: [12,60,60]

Explanation:

In the first query, nums[0..2] has 6 subarrays [2], [8], [4], [2, 8], [8, 4], and [2, 8, 4] each with a respective XOR score of 2, 8, 4, 10, 12, and 6. The answer for the query is 12, the largest of all XOR scores.

In the second query, the subarray of nums[1..4] with the largest XOR score is nums[1..4] with a score of 60.

In the third query, the subarray of nums[0..5] with the largest XOR score is nums[1..4] with a score of 60.

Example 2:

Input: nums = [0,7,3,2,8,5,1], queries = [[0,3],[1,5],[2,4],[2,6],[5,6]]

Output: [7,14,11,14,5]

Explanation:

Index	nums[l_i..r_i]	Maximum XOR Score Subarray	Maximum Subarray XOR Score
0	[0, 7, 3, 2]	[7]	7
1	[7, 3, 2, 8, 5]	[7, 3, 2, 8]	14
2	[3, 2, 8]	[3, 2, 8]	11
3	[3, 2, 8, 5, 1]	[2, 8, 5, 1]	14
4	[5, 1]	[5]	5

Constraints:

1 <= n == nums.length <= 2000
0 <= nums[i] <= 2³¹ - 1
1 <= q == queries.length <= 10⁵
queries[i].length == 2
queries[i] = [l_i, r_i]
0 <= l_i <= r_i <= n - 1

Step 01

Brute Force Baseline

Problem summary: You are given an array nums of n integers, and a 2D integer array queries of size q, where queries[i] = [li, ri]. For each query, you must find the maximum XOR score of any subarray of nums[li..ri]. The XOR score of an array a is found by repeatedly applying the following operations on a so that only one element remains, that is the score: Simultaneously replace a[i] with a[i] XOR a[i + 1] for all indices i except the last one. Remove the last element of a. Return an array answer of size q where answer[i] is the answer to query i.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[2,8,4,32,16,1]
[[0,2],[1,4],[0,5]]

Example 2

[0,7,3,2,8,5,1]
[[0,3],[1,5],[2,4],[2,6],[5,6]]

Core Insight

What unlocks the optimal approach

Precompute the XOR score of every subarray.
Try to find a relationship between XOR score of <code>nums[i..j], nums[i..j + 1], nums[i..j + 2], …</code>. Do you notice any pattern?
If <code>dp[i][j]</code> is the XOR score of subarray <code>nums[i..j]</code>, <code>dp[i][j] = dp[i - 1][j] XOR dp[i - 1][j + 1]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
class Solution {
    public int[] maximumSubarrayXor(int[] nums, int[][] queries) {
        int n = nums.length;
        int[][] f = new int[n][n];
        int[][] g = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            f[i][i] = nums[i];
            g[i][i] = nums[i];
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = f[i][j - 1] ^ f[i + 1][j];
                g[i][j] = Math.max(f[i][j], Math.max(g[i][j - 1], g[i + 1][j]));
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            int l = queries[i][0], r = queries[i][1];
            ans[i] = g[l][r];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
func maximumSubarrayXor(nums []int, queries [][]int) (ans []int) {
	n := len(nums)
	f := make([][]int, n)
	g := make([][]int, n)
	for i := 0; i < n; i++ {
		f[i] = make([]int, n)
		g[i] = make([]int, n)
	}
	for i := n - 1; i >= 0; i-- {
		f[i][i] = nums[i]
		g[i][i] = nums[i]
		for j := i + 1; j < n; j++ {
			f[i][j] = f[i][j-1] ^ f[i+1][j]
			g[i][j] = max(f[i][j], max(g[i][j-1], g[i+1][j]))
		}
	}
	for _, q := range queries {
		l, r := q[0], q[1]
		ans = append(ans, g[l][r])
	}
	return
}

# Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
class Solution:
    def maximumSubarrayXor(
        self, nums: List[int], queries: List[List[int]]
    ) -> List[int]:
        n = len(nums)
        f = [[0] * n for _ in range(n)]
        g = [[0] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            f[i][i] = g[i][i] = nums[i]
            for j in range(i + 1, n):
                f[i][j] = f[i][j - 1] ^ f[i + 1][j]
                g[i][j] = max(f[i][j], g[i][j - 1], g[i + 1][j])
        return [g[l][r] for l, r in queries]

// Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
// class Solution {
//     public int[] maximumSubarrayXor(int[] nums, int[][] queries) {
//         int n = nums.length;
//         int[][] f = new int[n][n];
//         int[][] g = new int[n][n];
//         for (int i = n - 1; i >= 0; --i) {
//             f[i][i] = nums[i];
//             g[i][i] = nums[i];
//             for (int j = i + 1; j < n; ++j) {
//                 f[i][j] = f[i][j - 1] ^ f[i + 1][j];
//                 g[i][j] = Math.max(f[i][j], Math.max(g[i][j - 1], g[i + 1][j]));
//             }
//         }
//         int m = queries.length;
//         int[] ans = new int[m];
//         for (int i = 0; i < m; ++i) {
//             int l = queries[i][0], r = queries[i][1];
//             ans[i] = g[l][r];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3277: Maximum XOR Score Subarray Queries
function maximumSubarrayXor(nums: number[], queries: number[][]): number[] {
    const n = nums.length;
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    const g: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = n - 1; i >= 0; i--) {
        f[i][i] = nums[i];
        g[i][i] = nums[i];
        for (let j = i + 1; j < n; j++) {
            f[i][j] = f[i][j - 1] ^ f[i + 1][j];
            g[i][j] = Math.max(f[i][j], Math.max(g[i][j - 1], g[i + 1][j]));
        }
    }
    return queries.map(([l, r]) => g[l][r]);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2 + m)

Space

O(n^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.