#1035

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Uncrossed Lines

The Uncrossed Lines problem involves finding the maximum number of non-intersecting lines that can be drawn between two integer arrays using dynamic programming.

array dynamic programming

Problem Statement

You are given two integer arrays, nums1 and nums2. Write the integers of nums1 and nums2 on two horizontal lines. You can draw a connecting line between nums1[i] and nums2[j] if nums1[i] == nums2[j]. However, no two lines can intersect, meaning that if a line is drawn between nums1[i] and nums2[j], the indices of nums1 and nums2 must increase for subsequent lines.

The task is to find the maximum number of uncrossed lines that can be drawn between nums1 and nums2. You must consider all possible ways to draw the lines, and the challenge is to use dynamic programming to solve this problem optimally.

Examples

Example 1

Input: nums1 = [1,4,2], nums2 = [1,2,4]

Output: 2

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 details omitted.

Example 3

Input: nums1 = [1,3,7,1,7,5], nums2 = [1,9,2,5,1]

Output: 2

Example details omitted.

Constraints

1 <= nums1.length, nums2.length <= 500
1 <= nums1[i], nums2[j] <= 2000

Solution Approach

State Transition Dynamic Programming

This problem can be solved using a dynamic programming approach where dp[i][j] represents the maximum number of uncrossed lines that can be drawn using the subarrays nums1[i:] and nums2[j:]. If nums1[i] == nums2[j], then dp[i][j] = dp[i-1][j-1] + 1, otherwise dp[i][j] = max(dp[i-1][j], dp[i][j-1]).

Base Case Initialization

The base case of the dynamic programming table is when either nums1 or nums2 is empty. In such cases, no lines can be drawn, so dp[i][0] = 0 for all i, and dp[0][j] = 0 for all j.

Time and Space Optimization

The space complexity can be optimized from O(n1 * n2) to O(n2) by only storing the current and previous rows of the DP table, since each cell only depends on the previous row and the current row.

Complexity Analysis

Metric	Value
Time	O(n1 \cdot n2)
Space	O(n2)

The time complexity is O(n1 * n2) where n1 and n2 are the lengths of nums1 and nums2 respectively. This is due to filling the dynamic programming table which has dimensions n1 x n2. The space complexity can be reduced to O(n2) by optimizing the storage of the DP table.

What Interviewers Usually Probe

The candidate demonstrates an understanding of dynamic programming and its application to problems with overlapping subproblems.
The candidate efficiently uses space optimization techniques for dynamic programming.
The candidate can identify the base cases and correctly initialize the dynamic programming table.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the base cases when either nums1 or nums2 is empty.
Incorrectly filling the DP table, particularly when updating dp[i][j] based on conditions like nums1[i] == nums2[j].
Using too much space for the DP table without considering the optimization to reduce space complexity.

Follow-up variants

Implementing a solution that handles multiple arrays rather than just two.
Extending the problem to consider more than one match between elements at the same index.
Considering the problem with additional constraints, like limiting the number of allowed lines.

FAQ

How do I approach the Uncrossed Lines problem?

The Uncrossed Lines problem can be solved using dynamic programming. Focus on using state transitions to keep track of the best solutions for subarrays of nums1 and nums2.

What is the time complexity of the Uncrossed Lines problem?

The time complexity is O(n1 * n2), where n1 and n2 are the lengths of the input arrays nums1 and nums2, respectively.

Can I optimize the space complexity for the Uncrossed Lines problem?

Yes, you can optimize the space complexity to O(n2) by storing only the current and previous rows of the dynamic programming table.

What is the base case in dynamic programming for Uncrossed Lines?

The base case occurs when either nums1 or nums2 is empty, in which case no lines can be drawn, so dp[i][0] = 0 and dp[0][j] = 0.

How does dynamic programming help solve the Uncrossed Lines problem?

Dynamic programming helps by breaking the problem down into smaller subproblems and storing solutions to subarrays, avoiding redundant computations and efficiently finding the maximum number of uncrossed lines.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ to represent the maximum number of connections between the first $i$ numbers of $\textit{nums1}$ and the first $j$ numbers of $\textit{nums2}$. Initially, $f[i][j] = 0$, and the answer is $f[m][n]$.

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class Solution:
    def maxUncrossedLines(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i, x in enumerate(nums1, 1):
            for j, y in enumerate(nums2, 1):
                if x == y:
                    f[i][j] = f[i - 1][j - 1] + 1
                else:
                    f[i][j] = max(f[i - 1][j], f[i][j - 1])
        return f[m][n]

Continue Practicing

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