Introduction

In this blog post, we will explore how to solve the LeetCode problem "3288" 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

You are given a 2D array of integers coordinates of length n and an integer k, where 0 <= k < n. coordinates[i] = [xi, yi] indicates the point (xi, yi) in a 2D plane. An increasing path of length m is defined as a list of points (x1, y1), (x2, y2), (x3, y3), ..., (xm, ym) such that: xi < xi + 1 and yi < yi + 1 for all i where 1 <= i < m. (xi, yi) is in the given coordinates for all i where 1 <= i <= m. Return the maximum length of an increasing path that contains coordinates[k]. Example 1: Input: coordinates = [[3,1],[2,2],[4,1],[0,0],[5,3]], k = 1 Output: 3 Explanation: (0, 0), (2, 2), (5, 3) is the longest increasing path that contains (2, 2). Example 2: Input: coordinates = [[2,1],[7,0],[5,6]], k = 2 Output: 2 Explanation: (2, 1), (5, 6) is the longest increasing path that contains (5, 6). Constraints: 1 <= n == coordinates.length <= 105 coordinates[i].length == 2 0 <= coordinates[i][0], coordinates[i][1] <= 109 All elements in coordinates are distinct. 0 <= k <= n - 1

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Key Idea: Use dynamic programming (memoization) to avoid recomputing the longest increasing path starting from each point. The DP state dp[i] represents the length of the longest increasing path starting from coordinates[i].

• DFS Traversal: Perform a Depth-First Search (DFS) from the k-th coordinate to find the longest increasing path that includes it. In the DFS, explore all possible next points that satisfy the increasing path condition (x_i < x_i+1 and y_i < y_i+1).

• Memoization: Store the result of the DFS for each point in the dp array. This way, if we encounter a point again during the search, we can simply return the stored result instead of recomputing it.

• Complexity:

  • Runtime: O(N²) in the worst case, but in practice closer to O(N log N) because of the x and y conditions that prune the search. O(N) space due to the size of the dp array in the worst case, where N is the number of coordinates.

Code

    def solve():
    def longest_increasing_path(coordinates, k):
        n = len(coordinates)
        dp = {}  # Memoization: dp[i] stores the length of the longest increasing path starting from coordinates[i]

        def dfs(i):
            if i in dp:
                return dp[i]

            max_len = 1  # Initialize with the path length starting from the current point itself
            for j in range(n):
                if coordinates[j][0] > coordinates[i][0] and coordinates[j][1] > coordinates[i][1]:
                    max_len = max(max_len, 1 + dfs(j))

            dp[i] = max_len
            return max_len

        return dfs(k)

    # Example Usage (replace with your actual input)
    coordinates = [[3, 1], [2, 2], [4, 1], [0, 0], [5, 3]]
    k = 1
    result = longest_increasing_path(coordinates, k)
    print(result)  # Output: 3

solve()