329.Longest_Increasing_Path_in_a_Matrix(Hard).md

329. Longest Increasing Path in a Matrix (Hard)

Date and Time: Dec 25, 2024, 23:46 (EST)

Link: https://leetcode.com/problems/longest-increasing-path-in-a-matrix

Question:

Given an m x n integers matrix, return the length of the longest increasing path in matrix.

From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).

Example 1:

Input: matrix = [[9,9,4],[6,6,8],[2,1,1]]

Output: 4

Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:

Input: matrix = [[3,4,5],[3,2,6],[2,2,1]]

Output: 4

Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Example 3:

Input: matrix = [[1]]

Output: 1

Constraints:

• m == matrix.length

• n == matrix[i].length

• 1 <= m, n <= 200

• 0 <= matrix[i][j] <= 2^31 - 1

Walk-through:

1. Build 2D DP as cache for each entry's Longest Increasing Path.

2. Run DFS from each entry to their neighbors to find their Longest Increasing Path, update dp[r][c] = 1 + max_path.

Python Solution:

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        # Build 2D DP to cache each entry's longest increasing path
        # Run DFS from each entry to find the entry's neighbors' LIP + 1

        # TC: O(m*n), m=rows, n=cols, SC: O(m*n)
        rows, cols = len(matrix), len(matrix[0])
        dp = [[0] * cols for _ in range(rows)]
        directions = [[-1, 0], [1, 0], [0, 1], [0, -1]]
        res = 0
        # Save the result into dp, and return dp[r][c]
        def dfs(r, c):
            if not dp[r][c]:
                max_path = 0
                for dr, dc in directions:
                    row, col = dr+r, dc+c
                    if row in range(rows) and col in range(cols) and matrix[row][col] > matrix[r][c]:
                        max_path = max(max_path, dfs(row, col))
                dp[r][c] = 1 + max_path
            return dp[r][c]

        for r in range(rows):
            for c in range(cols):
                res = max(res, dfs(r, c))
        return res

Time Complexity: $O(mn)$
Space Complexity: $O(mn)$