What is the main strategy to solve Length of the Longest Increasing Path efficiently?

The core approach is binary search over the valid path length combined with memoized DFS over filtered and sorted coordinates.

Do we need to check all coordinates when computing the increasing path?

No, only coordinates with both x and y less than the target point need to be considered to ensure valid increasing paths.

Can this problem be solved without sorting the coordinates?

Sorting simplifies subsequence computations and ensures DFS builds paths in order, but brute force without sorting is possible but inefficient.

How does memoization improve performance in this problem?

Memoization stores intermediate longest path lengths from each point, avoiding repeated DFS traversals over the same subpaths.

Is the pattern 'Binary search over the valid answer space' common in other path problems?

Yes, this pattern efficiently narrows potential maximum path lengths and is key to solving problems like this with large input arrays.

#3288

Hard

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LeetCode 题解工作台

最长上升路径的长度

给你一个长度为 n 的二维整数数组 coordinates 和一个整数 k ，其中 0 。 coordinates[i] = [x i , y i ] 表示二维平面里一个点 (x i , y i ) 。如果一个点序列 (x 1 , y 1 ) , (x 2 , y 2 ) , (x 3 , y 3…

数组二分查找排序

题目描述

给你一个长度为 n 的二维整数数组 coordinates 和一个整数 k ，其中 0 <= k < n 。

coordinates[i] = [x_i, y_i] 表示二维平面里一个点 (x_i, y_i) 。

如果一个点序列 (x₁, y₁), (x₂, y₂), (x₃, y₃), ..., (x_m, y_m) 满足以下条件，那么我们称它是一个长度为 m 的 上升序列 ：

对于所有满足 1 <= i < m 的 i 都有 x_i < x_{i + 1} 且 y_i < y_{i + 1} 。
对于所有 1 <= i <= m 的 i 对应的点 (x_i, y_i) 都在给定的坐标数组里。

请你返回包含坐标 coordinates[k] 的 最长上升路径 的长度。

示例 1：

输入：coordinates = [[3,1],[2,2],[4,1],[0,0],[5,3]], k = 1

输出：3

解释：

(0, 0) ，(2, 2) ，(5, 3) 是包含坐标 (2, 2) 的最长上升路径。

示例 2：

输入：coordinates = [[2,1],[7,0],[5,6]], k = 2

输出：2

解释：

(2, 1) ，(5, 6) 是包含坐标 (5, 6) 的最长上升路径。

提示：

1 <= n == coordinates.length <= 10⁵
coordinates[i].length == 2
0 <= coordinates[i][0], coordinates[i][1] <= 10⁹
coordinates 中的元素 互不相同 。
0 <= k <= n - 1

lightbulb

解题思路

方法一

1

2

speed

复杂度分析

指标	值
时间	complexity mainly depends on the binary search iterations multiplied by the DFS or DP over filtered points, which is faster than brute force. Space complexity comes from memoization storage and the temporary filtered list, typically O(n) in the worst case.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Pay attention to filtering coordinates based on both x and y to avoid invalid paths.
question_mark
Consider using binary search over the maximum path length to reduce brute-force attempts.
question_mark
Sorting the coordinates before DFS simplifies subsequence calculations and improves efficiency.

warning

常见陷阱

外企场景

error
Not filtering coordinates strictly less than coordinates[k] leads to invalid paths being counted.
error
Attempting DFS without memoization causes timeouts for large arrays.
error
Confusing increasing path criteria by only checking x or y instead of both.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Compute the longest increasing path without requiring inclusion of a specific point.
arrow_right_alt
Find the longest decreasing path instead, reversing the comparison conditions.
arrow_right_alt
Allow non-strictly increasing sequences and measure the longest weakly increasing path.

help

常见问题

外企场景

继续练习

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