Problem Description

You are given an array nums of n integers and a 2D integer array queries of size q, where each query is represented as queries[i] = [li, ri].

The problem asks you to find the maximum XOR score of any subarray within the range nums[li..ri] for each query.

The XOR score of an array is calculated through a specific process:

1. Start with an array a
2. Repeatedly apply these operations until only one element remains:
   • For all indices i except the last one, simultaneously replace a[i] with a[i] XOR a[i + 1]
   • Remove the last element of the array
3. The final remaining element is the XOR score

For example, if we have an array [a, b, c, d]:

• First operation: [a XOR b, b XOR c, c XOR d]
• Second operation: [(a XOR b) XOR (b XOR c), (b XOR c) XOR (c XOR d)]
• Third operation: [((a XOR b) XOR (b XOR c)) XOR ((b XOR c) XOR (c XOR d))]
• This final value is the XOR score

For each query [l, r], you need to:

1. Consider all possible subarrays within nums[l..r]
2. Calculate the XOR score for each subarray
3. Return the maximum XOR score among all these subarrays

The output should be an array answer of size q where answer[i] contains the maximum XOR score for query i.

Intuition

Let's first understand what happens when we compute the XOR score. If we trace through the operations for a subarray, we notice a pattern. For an array [a, b, c]:

• After first operation: [a XOR b, b XOR c]
• After second operation: [(a XOR b) XOR (b XOR c)] = [a XOR c]

This reveals that the XOR score follows a recursive pattern similar to Pascal's triangle, but with XOR operations instead of additions.

The key insight is that we can compute the XOR score of any subarray nums[i..j] using the XOR scores of its smaller subarrays. Specifically, if we know the XOR score of nums[i..j-1] and nums[i+1..j], we can compute the XOR score of nums[i..j] by XORing these two values:

f[i][j] = f[i][j-1] XOR f[i+1][j]

This works because:

• f[i][j-1] represents the result after applying the XOR operations on elements from index i to j-1
• f[i+1][j] represents the result after applying the XOR operations on elements from index i+1 to j
• When we XOR these two values, we get the result for the full range [i, j]

Now, for each query [l, r], we need the maximum XOR score among all subarrays within this range. Instead of recalculating for each query, we can precompute another DP table g[i][j] that stores the maximum XOR score for any subarray within nums[i..j].

The maximum value g[i][j] can be:

1. The XOR score of the current range itself: f[i][j]
2. The maximum from the range excluding the last element: g[i][j-1]
3. The maximum from the range excluding the first element: g[i+1][j]

Therefore: g[i][j] = max(f[i][j], g[i][j-1], g[i+1][j])

By precomputing both tables, we can answer each query in O(1) time by simply looking up g[l][r].

Learn more about Dynamic Programming patterns.

Solution Approach

We implement the solution using dynamic programming with two 2D tables:

1. Table f[i][j]: Stores the XOR score of the subarray nums[i..j]
2. Table g[i][j]: Stores the maximum XOR score among all subarrays within nums[i..j]

Step 1: Initialize the DP tables

Create two n × n matrices where n is the length of the input array:

f = [[0] * n for _ in range(n)]
g = [[0] * n for _ in range(n)]

Step 2: Fill the DP tables bottom-up

We iterate from the bottom-right to top-left of the matrix (i.e., i from n-1 to 0). For each starting position i, we consider all ending positions j from i to n-1:

for i in range(n - 1, -1, -1):
    f[i][i] = g[i][i] = nums[i]  # Base case: single element
    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])

The base case is when i == j, meaning we have a single element. In this case:

• The XOR score is the element itself: f[i][i] = nums[i]
• The maximum is also the element itself: g[i][i] = nums[i]

For larger subarrays where j > i:

• Calculate the XOR score using the recurrence relation: f[i][j] = f[i][j-1] XOR f[i+1][j]
• Calculate the maximum by taking the largest among:
  • Current XOR score: f[i][j]
  • Maximum in the range without the last element: g[i][j-1]
  • Maximum in the range without the first element: g[i+1][j]

Step 3: Answer the queries

Once both tables are populated, we can answer each query in constant time:

return [g[l][r] for l, r in queries]

For each query [l, r], we simply look up g[l][r] which contains the precomputed maximum XOR score for all subarrays within nums[l..r].

Time Complexity: O(n²) for preprocessing + O(q) for answering queries, where n is the length of the array and q is the number of queries.

Space Complexity: O(n²) for storing the two DP tables.

Example Walkthrough

Let's walk through the solution with a small example to understand how the DP tables are built.

Input: nums = [3, 8, 2], queries = [[0, 2]]

We'll build two tables:

• f[i][j]: XOR score of subarray from index i to j
• g[i][j]: Maximum XOR score among all subarrays within range [i, j]

Step 1: Initialize 3×3 tables

f = [[0, 0, 0],    g = [[0, 0, 0],
     [0, 0, 0],         [0, 0, 0],
     [0, 0, 0]]         [0, 0, 0]]

Step 2: Fill tables bottom-up (i from 2 to 0)

When i = 2:

• Base case: f[2][2] = g[2][2] = nums[2] = 2

When i = 1:

• Base case: f[1][1] = g[1][1] = nums[1] = 8
• For j = 2:
  • f[1][2] = f[1][1] ^ f[2][2] = 8 ^ 2 = 10
  • g[1][2] = max(f[1][2], g[1][1], g[2][2]) = max(10, 8, 2) = 10

When i = 0:

• Base case: f[0][0] = g[0][0] = nums[0] = 3
• For j = 1:
  • f[0][1] = f[0][0] ^ f[1][1] = 3 ^ 8 = 11
  • g[0][1] = max(f[0][1], g[0][0], g[1][1]) = max(11, 3, 8) = 11
• For j = 2:
  • f[0][2] = f[0][1] ^ f[1][2] = 11 ^ 10 = 1
  • g[0][2] = max(f[0][2], g[0][1], g[1][2]) = max(1, 11, 10) = 11

Final tables:

f = [[3, 11, 1],    g = [[3, 11, 11],
     [0,  8, 10],        [0,  8, 10],
     [0,  0,  2]]        [0,  0,  2]]

Let's verify f[0][2] = 1 is correct:

• Original array: [3, 8, 2]
• First operation: [3^8, 8^2] = [11, 10]
• Second operation: [11^10] = [1] ✓

Step 3: Answer queries For query [0, 2]: return g[0][2] = 11

This means the maximum XOR score among all subarrays in nums[0..2] is 11, which comes from the subarray [3, 8] (as we calculated f[0][1] = 11).

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2 + m), where n is the length of the array nums and m is the length of the array queries.

• The nested loops iterate from i = n-1 down to 0 and for each i, j goes from i+1 to n-1. This creates approximately n^2/2 iterations, which is O(n^2).
• Inside the inner loop, each operation (XOR and max comparisons) takes O(1) time.
• The list comprehension at the end processes m queries, each taking O(1) time to access the precomputed value in g[l][r], resulting in O(m) time.
• Total time complexity: O(n^2) + O(m) = O(n^2 + m).

The space complexity is O(n^2).

• Two 2D arrays f and g are created, each of size n × n, requiring O(n^2) space each.
• The output list stores m values, requiring O(m) space, but this is typically not counted as auxiliary space since it's the required output.
• Total space complexity: O(n^2) + O(n^2) = O(n^2).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect DP Table Iteration Order

One of the most common mistakes is iterating through the DP tables in the wrong order. The algorithm requires building the solution from smaller subarrays to larger ones, but the dependency pattern is not immediately obvious.

Wrong approach:

# This won't work - iterating in the wrong order
for i in range(n):
    for j in range(i, n):
        if i == j:
            dp_xor[i][j] = nums[i]
        else:
            dp_xor[i][j] = dp_xor[i][j-1] ^ dp_xor[i+1][j]  # i+1 might not be computed yet!

Why it fails: When computing dp_xor[i][j], we need both dp_xor[i][j-1] and dp_xor[i+1][j] to be already computed. If we iterate with i from 0 to n-1, dp_xor[i+1][j] won't be available yet.

Correct approach:

# Iterate from bottom-right to top-left
for i in range(n - 1, -1, -1):  # Start from the last row, move upward
    for j in range(i, n):        # For each row, move left to right
        # Now dp_xor[i+1][j] is guaranteed to be computed

2. Confusing XOR Score Calculation with Simple XOR

Another pitfall is misunderstanding what the "XOR score" means. It's not just XORing all elements in the subarray.

Wrong interpretation:

# This is NOT the XOR score - it's just a simple XOR of all elements
xor_score = nums[i]
for k in range(i + 1, j + 1):
    xor_score ^= nums[k]

Correct understanding: The XOR score is the result of a triangular reduction process where we repeatedly XOR adjacent elements and reduce the array size until one element remains. This forms a pattern similar to Pascal's triangle but with XOR operations.

3. Not Handling Base Cases Properly

Forgetting to initialize single-element subarrays can lead to incorrect results.

Missing base case:

for i in range(n - 1, -1, -1):
    for j in range(i + 1, n):  # Skips the case when i == j
        dp_xor[i][j] = dp_xor[i][j-1] ^ dp_xor[i+1][j]

Correct initialization:

for i in range(n - 1, -1, -1):
    dp_xor[i][i] = nums[i]      # Initialize base case first
    max_xor[i][i] = nums[i]
    for j in range(i + 1, n):    # Then handle larger subarrays
        # ...

4. Incorrect Maximum Calculation

When computing max_xor[i][j], forgetting to consider all three possibilities can lead to wrong answers.

Incomplete maximum calculation:

# Only considering current XOR score and left subrange
max_xor[i][j] = max(dp_xor[i][j], max_xor[i][j-1])
# Missing max_xor[i+1][j]!

Complete solution:

max_xor[i][j] = max(
    dp_xor[i][j],      # Current subarray's XOR score
    max_xor[i][j-1],   # Maximum in left subrange
    max_xor[i+1][j]    # Maximum in right subrange
)

This ensures we consider all possible subarrays within the range [i, j].