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:
class Solution: def getNeighbours(self, i, j, m, n): for x, y in [(i - 1, j), (i, j + 1), (i + 1, j), (i, j - 1)]: if 0 <= x < m and 0 <= y < n: yield (x, y) def longestTill(self, matrix, i, j, m, n): if (i, j) in self.cache: return self.cache[(i, j)] mlen = 1 for x, y in self.getNeighbours(i, j, m, n): if matrix[x][y] > matrix[i][j]: curr = 1 + self.longestTill(matrix, x, y, m, n) mlen = max(mlen, curr) self.cache[(i, j)] = mlen return mlen def longestIncreasingPath(self, matrix: List[List[int]]) -> int: m = len(matrix) n = len(matrix[0]) mlen = 1 self.cache = {} for i in range(m): for j in range(n): if (i, j) not in self.cache: curr = self.longestTill(matrix, i, j, m, n) mlen = max(mlen, curr) return mlen 
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