Introduction

In this blog post, we will explore how to solve the LeetCode problem "329" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

• DFS with Memoization: Perform Depth-First Search (DFS) from each cell in the matrix. To avoid redundant calculations, memoize the length of the longest increasing path starting from each cell.
  • Explore Four Directions: During DFS, explore the four possible directions (up, down, left, right) from the current cell. Only move to a neighboring cell if its value is greater than the current cell's value.
  • Return Maximum Length: For each cell, return the maximum length of the increasing path starting from that cell. The overall result is the maximum of these lengths across all cells.

• Runtime Complexity: O(m * n), where m and n are the dimensions of the matrix.
• Storage Complexity: O(m * n) due to the memoization table.

Code

    def longestIncreasingPath(matrix):
    """
    Finds the length of the longest increasing path in a matrix.

    Args:
        matrix: A list of lists of integers representing the matrix.

    Returns:
        The length of the longest increasing path.
    """

    if not matrix:
        return 0

    m, n = len(matrix), len(matrix[0])
    dp = [[0] * n for _ in range(m)]  # Memoization table

    def dfs(i, j):
        if dp[i][j] != 0:
            return dp[i][j]

        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        max_length = 1  # Minimum length is 1 (the cell itself)

        for dx, dy in directions:
            x, y = i + dx, j + dy
            if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]:
                length = 1 + dfs(x, y)
                max_length = max(max_length, length)

        dp[i][j] = max_length
        return max_length

    longest_path = 0
    for i in range(m):
        for j in range(n):
            longest_path = max(longest_path, dfs(i, j))

    return longest_path