How do you solve Uncrossed Lines on LeetCode?

Uncrossed Lines (LeetCode #1035) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: The problem can be visualized as finding the longest common subsequence between two arrays.

What is the brute force solution for Uncrossed Lines?

The brute force approach for Uncrossed Lines is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible combination of pairs from nums1 and nums2 to see if they can form uncrossed lines. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Uncrossed Lines?

Uncrossed Lines uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Longest Common Subsequence.

What are the interview tips for Uncrossed Lines?

Always clarify the problem statement and constraints before jumping into coding. Discuss your thought process and approach with the interviewer to show your understanding. Practice writing out the DP table on paper to visualize the relationships between subproblems.

What are common mistakes when solving Uncrossed Lines?

Students often forget to handle the base case in dynamic programming, leading to incorrect results. Not properly initializing the DP table can cause index errors or incorrect calculations.

#1035

Uncrossed Lines

Medium

ArrayDynamic ProgrammingDynamic ProgrammingLongest Common Subsequence

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m * n)

💡

Intuition

Time O(m * n)Space O(m * n)

The optimal solution uses dynamic programming to build a table that keeps track of the maximum number of uncrossed lines that can be formed up to each index of nums1 and nums2. This approach is efficient and avoids redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D DP array where dp[i][j] represents the maximum number of uncrossed lines that can be formed using nums1[0..i-1] and nums2[0..j-1].
2Step 2: Iterate through each element of nums1 and nums2. If nums1[i-1] equals nums2[j-1], set dp[i][j] = dp[i-1][j-1] + 1. Otherwise, set dp[i][j] = max(dp[i-1][j], dp[i][j-1]).
3Step 3: The answer will be in dp[nums1.length][nums2.length].

solution.py15 lines

1# Full working Python code
2
3def maxUncrossedLines(nums1, nums2):
4    m, n = len(nums1), len(nums2)
5    dp = [[0] * (n + 1) for _ in range(m + 1)]
6    for i in range(1, m + 1):
7        for j in range(1, n + 1):
8            if nums1[i - 1] == nums2[j - 1]:
9                dp[i][j] = dp[i - 1][j - 1] + 1
10            else:
11                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
12    return dp[m][n]
13
14# Example usage
15print(maxUncrossedLines([1,4,2], [1,2,4]))  # Output: 2

ℹ

Complexity note: The time complexity is O(m * n) because we are filling a 2D array of size m x n, where m and n are the lengths of nums1 and nums2, respectively. The space complexity is also O(m * n) due to the storage of the DP table.

1The problem can be visualized as finding the longest common subsequence between two arrays.
2Dynamic programming is a powerful technique for optimizing problems that involve making decisions based on previous results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.