3 ms·

It solves easy problems and gets nowhere close on hard problems. But I'm sure that's coming.

sayogo1227 4y ago
Here is what I tried for (https://leetcode.com/problems/longest-increasing-path-in-a-matrix/ https://leetcode.com/problems/longest-increasing-path-in-a-m...) INPUT: 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). Input: matrix = [[9,9,4],[6,6,8],[2,1,1]] Output: 4 Explanation: The longest increasing path is [1, 2, 6, 9]. class Solution: def longestIncreasingPath(self, matrix: List[List[int]]) -> int: ChatGPT: if not matrix or not matrix[0]: return 0 # initialize the 2D array with all zeros dp = [[0] * len(matrix[0]) for _ in range(len(matrix))] # initialize the result with 0 result = 0 # iterate through each cell in the matrix for i in range(len(matrix)): for j in range(len(matrix[0])): # perform DFS starting from the current cell result = max(result, self.dfs(matrix, dp, i, j)) # return the result return result # helper function to perform DFS def dfs(self, matrix, dp, i, j): # if the path length is already calculated, return it if dp[i][j] > 0: return dp[i][j] # initialize the path length with 1 path_len = 1 # check the adjacent cells for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]: # get the coordinates of the adjacent cell x, y = i + dx, j + dy # check if the adjacent cell is valid if 0 <= x < len(matrix) and 0 <= y < len(matrix[0]): # check if the adjacent cell has a smaller value if matrix[x][y] > matrix[i][j]: # perform DFS on the adjacent cell and update the path length path_len = max(path_len, 1 + self.dfs(matrix, dp, x, y)) # store the path length in the 2D array dp[i][j] = path_len # return the path length return path_len
- DenisM 4y ago
  Have you tried submitting it?
  - sayogo1227 4y ago
    Yes. Accepted.
    - DenisM 4y ago
      So what do we make of that? A really good search engine, or an entity that can synthesize?

Here is what I tried for (https://leetcode.com/problems/longest-increasing-path-in-a-matrix/ https://leetcode.com/problems/longest-increasing-path-in-a-m...) INPUT: 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). Input: matrix = [[9,9,4],[6,6,8],[2,1,1]] Output: 4 Explanation: The longest increasing path is [1, 2, 6, 9]. class Solution: def longestIncreasingPath(self, matrix: List[List[int]]) -> int: ChatGPT: if not matrix or not matrix[0]: return 0 # initialize the 2D array with all zeros dp = [[0] * len(matrix[0]) for _ in range(len(matrix))] # initialize the result with 0 result = 0 # iterate through each cell in the matrix for i in range(len(matrix)): for j in range(len(matrix[0])): # perform DFS starting from the current cell result = max(result, self.dfs(matrix, dp, i, j)) # return the result return result # helper function to perform DFS def dfs(self, matrix, dp, i, j): # if the path length is already calculated, return it if dp[i][j] > 0: return dp[i][j] # initialize the path length with 1 path_len = 1 # check the adjacent cells for dx, dy in [(1, 0), (-1, 0), (0, 1), (0, -1)]: # get the coordinates of the adjacent cell x, y = i + dx, j + dy # check if the adjacent cell is valid if 0 <= x < len(matrix) and 0 <= y < len(matrix[0]): # check if the adjacent cell has a smaller value if matrix[x][y] > matrix[i][j]: # perform DFS on the adjacent cell and update the path length path_len = max(path_len, 1 + self.dfs(matrix, dp, x, y)) # store the path length in the 2D array dp[i][j] = path_len # return the path length return path_len

DenisM 4y ago
Have you tried submitting it?
- sayogo1227 4y ago
  Yes. Accepted.
  - DenisM 4y ago
    So what do we make of that? A really good search engine, or an entity that can synthesize?