Introduction

In this blog post, we will explore how to solve the LeetCode problem "2713" 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 a 1-indexed m x n integer matrix mat, you can select any cell in the matrix as your starting cell. From the starting cell, you can move to any other cell in the same row or column, but only if the value of the destination cell is strictly greater than the value of the current cell. You can repeat this process as many times as possible, moving from cell to cell until you can no longer make any moves. Your task is to find the maximum number of cells that you can visit in the matrix by starting from some cell. Return an integer denoting the maximum number of cells that can be visited. Example 1: Input: mat = [[3,1],[3,4]] Output: 2 Explanation: The image shows how we can visit 2 cells starting from row 1, column 2. It can be shown that we cannot visit more than 2 cells no matter where we start from, so the answer is 2. Example 2: Input: mat = [[1,1],[1,1]] Output: 1 Explanation: Since the cells must be strictly increasing, we can only visit one cell in this example. Example 3: Input: mat = [[3,1,6],[-9,5,7]] Output: 4 Explanation: The image above shows how we can visit 4 cells starting from row 2, column 1. It can be shown that we cannot visit more than 4 cells no matter where we start from, so the answer is 4. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 105 1 <= m * n <= 105 -105 <= mat[i][j] <= 105

Explanation

Here's the approach, complexity analysis, and Python code for solving this problem:

• Key Idea: Use Dynamic Programming (DP) with Memoization. For each cell, the maximum path length starting from that cell can be calculated recursively. The base case is when no move is possible, which returns 1. Memoization avoids redundant calculations and optimizes the process.

• Monotonic Stack/Sorting: To optimize searching for the next valid cell to move to in the same row or column, sort cells by their values. Iterate through the sorted cells. For each cell, check its neighbors (cells in the same row and column). Use the DP table to find maximum path from neighbor cells, add +1 and update DP table value of current cell.

• Complexity:

  • Runtime: O(m*n*log(m*n)), due to sorting the cells. The DP part takes O(m*n) in the worst case.
  • Storage: O(m*n) for the DP table and storing cell coordinates.

Code

    def maxIncreasingCells(mat):
    m, n = len(mat), len(mat[0])
    dp = {}
    cells = []
    for r in range(m):
        for c in range(n):
            cells.append((mat[r][c], r, c))

    cells.sort()
    max_path = 0

    for val, r, c in cells:
        curr_path = 1
        # Check row
        for j in range(n):
            if j != c and mat[r][j] > val:
                if (r, j) in dp:
                    curr_path = max(curr_path, dp[(r, j)] + 1)

        # Check column
        for i in range(m):
            if i != r and mat[i][c] > val:
                if (i, c) in dp:
                    curr_path = max(curr_path, dp[(i, c)] + 1)

        dp[(r, c)] = curr_path
        max_path = max(max_path, curr_path)

    return max_path