329.Longest Increasing Path in a Matrix

Problem description:

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] <= 231 - 1

Solution:

1. naive: check all the node and calculate
2. Top down DP

for each cell, memorize the max increasing path could go

class Solution:
    def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
        # dp top down
        # for each cell, memorize the max increasing path could go
        self.res = 0
        dp = defaultdict()
        def dfs(x, y, prev):
            if x < 0 or y < 0 or x >= len(matrix) or y >= len(matrix[0]) or matrix[x][y] <= prev:
                return 0
            if (x, y) in dp:
                return dp[(x,y)]
            
            path = 1+ max(dfs(x-1, y, matrix[x][y]), dfs(x+1, y, matrix[x][y]), dfs(x, y-1, matrix[x][y]), dfs(x, y+1, matrix[x][y]))
            
            self.res = max(self.res, path)
            dp[(x,y)] = path
            return path
        
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                dfs(i, j, float("-inf"))
        return self.res

time complexity: $O()$
space complexity: $O()$
reference:
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