How do you solve Maximum Number of Points From Grid Queries on LeetCode?

Maximum Number of Points From Grid Queries (LeetCode #2503) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n log(m * n) + k log k) time complexity and O(m * n) space complexity. Key insight: Sorting the cells allows us to efficiently determine which cells can be visited for each query.

What is the time complexity of Maximum Number of Points From Grid Queries?

The optimal solution for Maximum Number of Points From Grid Queries runs in O(m * n log(m * n) + k log k) time and uses O(m * n) space. The time complexity is O(m * n log(m * n)) for sorting the cells and O(k log k) for sorting the queries. The space complexity is O(m * n) for storing the visited cells.

What is the brute force solution for Maximum Number of Points From Grid Queries?

The brute force approach for Maximum Number of Points From Grid Queries is Brute Force, with O(n²) time complexity and O(m * n) space complexity. In the brute force approach, we will explore all possible paths from the top-left cell of the grid for each query. We will count points based on whether the current cell value is less than the query value, and we will stop when we encounter a cell that doesn't meet this condition. The optimal Optimal Solution approach improves this to O(m * n log(m * n) + k log k).

What are the interview tips for Maximum Number of Points From Grid Queries?

Always consider sorting as a strategy when dealing with multiple queries. Keep track of the original indices of queries when sorting to maintain the output order. Discuss your thought process and approach before jumping into coding; it helps clarify your understanding.

What are common mistakes when solving Maximum Number of Points From Grid Queries?

Failing to sort the queries and cells, leading to inefficient traversal. Not using a visited set, which can result in over-counting points.

#2503

Maximum Number of Points From Grid Queries

Hard

ArrayTwo PointersBreadth-First SearchUnion-FindSortingHeap (Priority Queue)MatrixSortingBreadth-First SearchGraph Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n log(m * n) + k log k)
Space	O(m * n)	O(m * n)

💡

Intuition

Time O(m * n log(m * n) + k log k)Space O(m * n)

In the optimal solution, we will sort the queries and use a priority queue to efficiently explore the grid. By processing the queries in ascending order, we can avoid redundant calculations and only explore cells that are relevant to the current query.

⚙️

Algorithm

3 steps

1Step 1: Create a list of all cells in the grid with their values and sort them.
2Step 2: Sort the queries and keep track of their original indices.
3Step 3: Use a priority queue to process cells in ascending order of their values, while keeping track of the number of points for each query based on the cells that have been processed.

solution.py28 lines

1# Full working Python code
2from collections import deque
3import heapq
4
5def maxPointsOptimal(grid, queries):
6    m, n = len(grid), len(grid[0])
7    cells = [(grid[i][j], i, j) for i in range(m) for j in range(n)]
8    cells.sort()
9    sorted_queries = sorted((q, i) for i, q in enumerate(queries))
10    results = [0] * len(queries)
11    visited = set()
12    directions = [(-1, 0), (1, 0), (0, -1), (0, 1)]
13    points = 0
14    queue = deque([(0, 0)])
15
16    for query, original_index in sorted_queries:
17        while queue and grid[0][0] < query:
18            x, y = queue.popleft()
19            if (x, y) in visited:
20                continue
21            visited.add((x, y))
22            points += 1
23            for dx, dy in directions:
24                nx, ny = x + dx, y + dy
25                if 0 <= nx < m and 0 <= ny < n and (nx, ny) not in visited and grid[nx][ny] < query:
26                    queue.append((nx, ny))
27        results[original_index] = points
28    return results

ℹ

Complexity note: The time complexity is O(m * n log(m * n)) for sorting the cells and O(k log k) for sorting the queries. The space complexity is O(m * n) for storing the visited cells.

1Sorting the cells allows us to efficiently determine which cells can be visited for each query.
2Using a visited set prevents counting the same cell multiple times, optimizing the point calculation.

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