Uncrossed Lines - Video Solutions

Problem Overview: You are given two integer arrays. Draw lines connecting equal numbers between the arrays such that the lines do not cross. The goal is to compute the maximum number of such connections. The key observation: once two elements are matched, their relative order must remain consistent, otherwise the lines would intersect.

This constraint makes the problem identical to the Longest Common Subsequence (LCS) problem. You want the longest sequence of elements that appears in both arrays while preserving order. Every element in that subsequence represents one uncrossed line.

Approach 1: Dynamic Programming (LCS Table) (Time: O(n*m), Space: O(n*m))

Create a 2D DP table where dp[i][j] represents the maximum number of uncrossed lines you can draw using the first i elements of nums1 and the first j elements of nums2. Iterate through both arrays. If nums1[i-1] == nums2[j-1], you can connect them, so dp[i][j] = 1 + dp[i-1][j-1]. Otherwise, skip one element from either array: dp[i][j] = max(dp[i-1][j], dp[i][j-1]). This builds the optimal answer bottom‑up. The approach systematically explores all pair combinations and ensures order is preserved, preventing crossings. The final answer is stored in dp[n][m].

This method is straightforward and commonly used in interview settings because it clearly demonstrates understanding of dynamic programming. The table explicitly shows how partial solutions combine to form the final result.

Approach 2: Space Optimized Dynamic Programming (Time: O(n*m), Space: O(min(n,m)))

The full DP table is not always necessary. Each state only depends on the previous row and the current row. Instead of storing an n x m matrix, maintain two 1D arrays representing the previous and current rows. Iterate through nums1, and for each element update values across nums2. When the elements match, update using the diagonal value from the previous row; otherwise carry forward the maximum from adjacent states.

This optimization reduces memory usage significantly while preserving the same O(n*m) runtime. It is useful when arrays are large and memory limits matter. The logic remains identical to the classic LCS DP, but storage shrinks to linear space.

The problem is categorized under Array and Dynamic Programming because it combines sequential array traversal with optimal substructure reasoning. Once you recognize the LCS pattern, the solution becomes straightforward.

Recommended for interviews: The standard 2D dynamic programming solution is what interviewers typically expect. It clearly models the LCS recurrence and demonstrates strong DP fundamentals. Mentioning the space optimization afterward shows deeper understanding and awareness of memory tradeoffs.