# Solving Leetcode Interviews in Seconds with AI: Length of the Longest Increasing Path


	# 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](https://www.chatmagic.app), 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<sub>i</sub> < x<sub>i+1</sub> and y<sub>i</sub> < y<sub>i+1</sub>).

*   **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(N<sup>2</sup>) 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
	```python
	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()
	```