How do you solve Maximum Sum Queries on LeetCode?

Maximum Sum Queries (LeetCode #2736) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + q log q) time complexity and O(n) space complexity. Key insight: Sorting the input arrays allows us to efficiently manage the constraints of the queries.

What is the brute force solution for Maximum Sum Queries?

The brute force approach for Maximum Sum Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check each index in the arrays for every query to find the maximum sum that meets the conditions. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n + q log q).

What are the interview tips for Maximum Sum Queries?

Always discuss your thought process and approach before jumping into coding. Consider edge cases, such as empty inputs or all elements being smaller than the query constraints. Practice explaining your code and the reasoning behind your choices, as this is crucial in interviews.

What are common mistakes when solving Maximum Sum Queries?

Not considering the case where no valid indices exist, leading to incorrect results. Failing to sort the queries and pairs properly, which can lead to inefficient solutions.

#2736

Maximum Sum Queries

Hard

ArrayBinary SearchStackBinary Indexed TreeSegment TreeSortingMonotonic StackSortingHeapTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + q log q)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + q log q)Space O(n)

By sorting the input arrays and using a data structure to efficiently track valid sums, we can significantly reduce the number of comparisons needed for each query.

⚙️

Algorithm

5 steps

1Step 1: Create a list of tuples (nums1[i], nums2[i]) and sort it by nums1[i] in descending order.
2Step 2: Create a list of queries with their original indices and sort them by x-coordinate in descending order.
3Step 3: Initialize a max-heap (or priority queue) to keep track of valid sums as we process each query.
4Step 4: For each query, while there are elements in the sorted tuples that meet the x condition, add their sums to the heap if they also meet the y condition.
5Step 5: For each query, the maximum sum can be retrieved from the heap; if empty, return -1.

solution.py16 lines

1import heapq
2
3def maxSumQueries(nums1, nums2, queries):
4    indexed_queries = sorted(enumerate(queries), key=lambda x: -x[1][0])
5    pairs = sorted(zip(nums1, nums2), key=lambda x: -x[0])
6    results = [-1] * len(queries)
7    max_heap = []
8    j = 0
9    for idx, (x, y) in indexed_queries:
10        while j < len(pairs) and pairs[j][0] >= x:
11            heapq.heappush(max_heap, pairs[j][0] + pairs[j][1])
12            j += 1
13        while max_heap and pairs[j-1][1] < y:
14            heapq.heappop(max_heap)
15        results[idx] = max_heap[0] if max_heap else -1
16    return results

ℹ

Complexity note: The time complexity is dominated by the sorting steps, which are O(n log n) for the pairs and O(q log q) for the queries. The space complexity is O(n) due to the storage of pairs.

1Sorting the input arrays allows us to efficiently manage the constraints of the queries.
2Using a max-heap helps in quickly retrieving the maximum sum that meets the conditions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.