Introduction

In this blog post, we will explore how to solve the LeetCode problem "1035" 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 two integer arrays nums1 and nums2. We write the integers of nums1 and nums2 (in the order they are given) on two separate horizontal lines. We may draw connecting lines: a straight line connecting two numbers nums1[i] and nums2[j] such that: nums1[i] == nums2[j], and the line we draw does not intersect any other connecting (non-horizontal) line. Note that a connecting line cannot intersect even at the endpoints (i.e., each number can only belong to one connecting line). Return the maximum number of connecting lines we can draw in this way. Example 1: Input: nums1 = [1,4,2], nums2 = [1,2,4] Output: 2 Explanation: We can draw 2 uncrossed lines as in the diagram. We cannot draw 3 uncrossed lines, because the line from nums1[1] = 4 to nums2[2] = 4 will intersect the line from nums1[2]=2 to nums2[1]=2. Example 2: Input: nums1 = [2,5,1,2,5], nums2 = [10,5,2,1,5,2] Output: 3 Example 3: Input: nums1 = [1,3,7,1,7,5], nums2 = [1,9,2,5,1] Output: 2 Constraints: 1 <= nums1.length, nums2.length <= 500 1 <= nums1[i], nums2[j] <= 2000

Explanation

Here's the solution to the problem:

• Key Idea: The problem can be modeled as finding the Longest Common Subsequence (LCS) between the two arrays. Each common element represents a potential connecting line, and the non-intersecting constraint translates to maintaining the order of elements in both arrays, which is precisely what LCS finds.

• Dynamic Programming: We use dynamic programming to compute the LCS. dp[i][j] stores the length of the LCS of nums1[:i] and nums2[:j].

• Optimal Substructure: If nums1[i-1] == nums2[j-1], then dp[i][j] = dp[i-1][j-1] + 1. Otherwise, dp[i][j] = max(dp[i-1][j], dp[i][j-1]).

• Runtime & Storage Complexity: O(m*n) where m and n are the lengths of nums1 and nums2 respectively.

Code

    def maxUncrossedLines(nums1, nums2):
    """
    Finds the maximum number of uncrossed lines between two arrays.

    Args:
        nums1: The first integer array.
        nums2: The second integer array.

    Returns:
        The maximum number of uncrossed lines.
    """

    m = len(nums1)
    n = len(nums2)

    # Initialize a DP table to store the lengths of LCS
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    # Iterate through the arrays to populate the DP table
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if nums1[i - 1] == nums2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    # The result is the length of the LCS of the entire arrays
    return dp[m][n]