Longest Increasing Path in a Matrix
Created: April 6, 2020 by [lek-tin]
Last updated: April 6, 2020
Given an integer matrix, find the length of the longest increasing path.
From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).
Example 1
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2
Input: nums =
[
[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.
Solution
class Solution:
def longestIncreasingPath(self, matrix: List[List[int]]) -> int:
if not matrix or not matrix[0]:
return 0
self.N, self.M = len(matrix), len(matrix[0])
cache = [ [0 for j in range(self.M)] for i in range(self.N) ]
longest = 1
for i in range(self.N):
for j in range(self.M):
res = self.dfs(matrix, i, j, cache)
if longest < res:
longest = res
return longest
def dfs(self, matrix, i, j, cache):
if cache[i][j] != 0:
return cache[i][j]
dirs = [ [1,0], [-1,0], [0, 1], [0, -1] ]
longest = 1
for dx, dy in dirs:
next_i, next_j = i+dx, j+dy
# if matrix[i][j] < matrix[next_i][next_j], then it is not possible that
# our dfs func previously walked from (next_i, next_j) ->(i, j)
if 0 <= next_i < self.N and 0 <= next_j < self.M and matrix[i][j] < matrix[next_i][next_j]:
next_l = 1 + self.dfs(matrix, next_i, next_j, cache)
if longest < next_l:
longest = next_l
cache[i][j] = longest
return longest