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LinkedIn·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026Remote

Summary

LinkedIn SWE coding round, pretty focused on DP. One problem, one optimization discussion, and then it was over. Left feeling like I could've explained the O(n*k) trick more cleanly.

Questions Asked (1)

Q1

Given n houses in a row and k colors, where costs[i][j] is the cost to paint house i with color j and no two adjacent houses can share the same color, find the minimum total cost to paint all houses.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I got the DP setup pretty fast.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain a dynamic programming solution where dp[i][j] represents the minimum cost to paint the first i houses with house i painted color j. Optimize the transition by tracking the two smallest costs from the previous row, reducing time complexity from O(n*k^2) to O(n*k).

Pro tip: Mention that you can optimize space to O(k) by only keeping the previous row's costs, and highlight that this problem is a classic example of dynamic programming with optimization, often asked at top tech companies.

1. Clarify and Restate

Confirm the problem details: n houses, k colors, costs matrix, and adjacency constraint. Ask about edge cases like n=0 or k=1.

2. Define DP State

Define dp[i][j] as the minimum cost to paint houses 0..i with house i painted color j. Base case: dp[0][j] = costs[0][j].

3. Derive Transition

For each house i and color j, dp[i][j] = costs[i][j] + min(dp[i-1][c]) for all c != j. Naively this is O(n*k^2).

4. Optimize Transition

Track the smallest and second smallest values from the previous row along with their color indices. Then for each j, the min excluding j is either the smallest (if j != index of smallest) or the second smallest.

5. Analyze Complexity and Edge Cases

Time complexity: O(n*k), space: O(k) with rolling array. Handle n=0 (return 0), k=1 (if n>1, impossible).

Key Points to Mention

  • Dynamic programming state definition and transition
  • Optimization using two smallest costs to achieve O(n*k) time
  • Space optimization to O(k) using rolling array
  • Edge cases: n=0, k=1, large n and k
  • Comparison with naive O(n*k^2) approach and why optimization matters
  • Potential follow-up: what if k is very large? (still O(n*k) is optimal)

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.