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Uber·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Uber SWE interview with a classic dynamic programming string problem. Not a lot of context to go on but the question itself is a solid one that trips people up if they haven't seen it before.

Questions Asked (1)

Q1

Given a string, find the minimum number of characters you need to insert to make it a palindrome.

Algorithms & Data Structures
Author's notes

This is a DP problem but it's easy to go down the wrong path if you start thinking about it greedily.

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AI HintsAI Generated

Suggested Approach

Recognize that the minimum insertions equal the string length minus the length of the longest palindromic subsequence (LPS). Compute LPS using dynamic programming (or by finding the longest common subsequence between the string and its reverse). Then return n - LPS length.

Pro tip: After presenting the DP solution, mention that the problem can also be solved with a 2D DP where dp[i][j] = min insertions for substring s[i..j], and that this approach directly yields the answer without the LPS detour. This shows deeper understanding and flexibility.

1. Clarify and Restate

Confirm that insertions can be at any position and that we want the minimum number. Restate the problem in your own words to ensure alignment.

2. Identify Key Insight

Explain that the minimum insertions equal the number of characters not part of the longest palindromic subsequence. Thus, answer = n - LPS length.

3. Choose Computation Method

Describe how to compute LPS: either via DP on the string or by finding LCS between the string and its reverse. Mention time and space complexity (O(n^2)).

4. Walk Through Example

Pick a short example (e.g., 'ab') and show step-by-step how the LPS is found and the insertions calculated.

5. Discuss Optimizations and Edge Cases

Mention space optimization (e.g., using 1D DP for LCS) and handle edge cases like empty string or already palindrome.

Key Points to Mention

  • Longest palindromic subsequence (LPS) and its relation to minimum insertions
  • Dynamic programming recurrence for LPS: if s[i]==s[j], dp[i][j]=dp[i+1][j-1]+2; else max(dp[i+1][j], dp[i][j-1])
  • Alternative: LCS between string and its reverse
  • Time and space complexity: O(n^2) time, O(n^2) space (can be optimized to O(n) space)
  • Edge cases: empty string (0 insertions), single character (0), already palindrome (0)
  • Example walkthrough to illustrate the concept

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