What is the two-pointer scanning approach for solving Maximum Number of Points From Grid Queries?

The two-pointer scanning approach helps you efficiently traverse the grid by utilizing a pointer system that tracks visited cells and ensures points are maximized during traversal.

How does invariant tracking help in the Maximum Number of Points From Grid Queries problem?

Invariant tracking ensures that once a part of the grid has been processed for a query, it is not recalculated unnecessarily for subsequent queries, improving overall efficiency.

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

The time complexity of the solution is O(k \log k + (n \cdot m) \log (n \cdot m) + k \cdot \alpha(n \cdot m)), driven by sorting and efficient grid traversal.

What are the common pitfalls when solving the Maximum Number of Points From Grid Queries problem?

Common pitfalls include not precomputing queries, mishandling grid traversal states, and failing to optimize query processing, leading to inefficiencies.

How can GhostInterview help solve the Maximum Number of Points From Grid Queries problem?

GhostInterview’s solver aids in applying two-pointer strategies, tracks traversal invariants, and guides you in optimizing query handling for grid problems.

#2503

Hard

auto_awesome双·指针·invariant

LeetCode 题解工作台

矩阵查询可获得的最大分数

给你一个大小为 m x n 的整数矩阵 grid 和一个大小为 k 的数组 queries 。找出一个大小为 k 的数组 answer ，且满足对于每个整数 queries[i] ，你从矩阵左上角单元格开始，重复以下过程：如果 queries[i] 严格大于你当前所处位置单元格，如果该单元…

数组双指针广度优先搜索并查集排序

题目描述

给你一个大小为 m x n 的整数矩阵 grid 和一个大小为 k 的数组 queries 。

找出一个大小为 k 的数组 answer ，且满足对于每个整数 queries[i] ，你从矩阵 左上角 单元格开始，重复以下过程：

如果 queries[i] 严格大于你当前所处位置单元格，如果该单元格是第一次访问，则获得 1 分，并且你可以移动到所有 4 个方向（上、下、左、右）上任一相邻单元格。
否则，你不能获得任何分，并且结束这一过程。

在过程结束后，answer[i] 是你可以获得的最大分数。注意，对于每个查询，你可以访问同一个单元格多次。

返回结果数组 answer 。

示例 1：

输入：grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]
输出：[5,8,1]
解释：上图展示了每个查询中访问并获得分数的单元格。

示例 2：

输入：grid = [[5,2,1],[1,1,2]], queries = [3]
输出：[0]
解释：无法获得分数，因为左上角单元格的值大于等于 3 。

提示：

m == grid.length
n == grid[i].length
2 <= m, n <= 1000
4 <= m * n <= 10⁵
k == queries.length
1 <= k <= 10⁴
1 <= grid[i][j], queries[i] <= 10⁶

lightbulb

解题思路

方法一：离线查询 + BFS + 优先队列（小根堆）

根据题目描述我们知道，每个查询相互独立，查询的顺序不影响结果，并且题目要我们每次从左上角开始，统计所有可以访问的、且值小于当前查询值的单元格的个数。

因此，我们可以先对 queries 数组进行排序，然后从小到大依次处理每个查询。

我们用优先队列（小根堆）维护当前访问到的最小单元格的值，用数组或哈希表 vis 记录当前单元格是否已经访问过。初始时，将左上角单元格的数据 $(grid[0][0], 0, 0)$ 作为三元组加入优先队列，并将 vis[0][0] 置为 True。

对于每个查询 queries[i]，我们判断当前优先队列的最小值是否小于 queries[i]，如果是，则将当前最小值弹出，累加计数器 cnt，并将当前单元格的上下左右四个单元格加入优先队列，注意要判断是否已经访问过。重复上述操作，直到当前优先队列的最小值大于等于 queries[i]，此时 cnt 即为当前查询的答案。

时间复杂度 $O(k \times \log k + m \times n \log(m \times n))$ ，空间复杂度 $O(m \times n)$ 。其中 $m$ 和 $n$ 分别为网格的行数和列数，而 $k$ 为查询的个数。我们需要对 queries 数组进行排序，时间复杂度为 $O(k \times \log k)$ 。矩阵中的每个单元格最多只会被访问一次，每一次入队和出队的时间复杂度为 $O(\log(m \times n))$ 。因此，总时间复杂度为 $O(k \times \log k + m \times n \log(m \times n))$ 。

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class Solution:
    def maxPoints(self, grid: List[List[int]], queries: List[int]) -> List[int]:
        m, n = len(grid), len(grid[0])
        qs = sorted((v, i) for i, v in enumerate(queries))
        ans = [0] * len(qs)
        q = [(grid[0][0], 0, 0)]
        cnt = 0
        vis = [[False] * n for _ in range(m)]
        vis[0][0] = True
        for v, k in qs:
            while q and q[0][0] < v:
                _, i, j = heappop(q)
                cnt += 1
                for a, b in pairwise((-1, 0, 1, 0, -1)):
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and not vis[x][y]:
                        heappush(q, (grid[x][y], x, y))
                        vis[x][y] = True
            ans[k] = cnt
        return ans

speed

复杂度分析

指标	值
时间	O(k \log k + (n \cdot m) \log (n \cdot m) + k \cdot \alpha(n \cdot m))
空间	O((n \cdot m) + k)

psychology

面试官常问的追问

外企场景

question_mark
Tests understanding of two-pointer strategies in grid-based problems.
question_mark
Assesses the ability to manage grid traversal with invariant tracking and efficient data structures.
question_mark
Evaluates optimization skills when processing multiple queries in a grid setting.

warning

常见陷阱

外企场景

error
Not leveraging the precomputation of queries, which leads to redundant computations.
error
Incorrectly handling the grid traversal state, leading to errors in point calculation.
error
Inefficient query handling that doesn't scale well with large inputs or multiple queries.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider handling grids with larger dimensions to test the efficiency of the solution.
arrow_right_alt
Test with queries that are all smaller than the values in the grid to ensure the algorithm handles edge cases.
arrow_right_alt
Consider implementing a dynamic programming approach instead of sorting for different performance trade-offs.

help

常见问题

外企场景

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