Google Interview: RPG Combo Chain (O(N) Solution)

🎮 Background: RPG Combo Chain

In a real-time Google SDE interview assistance session, we encountered a deceptively simple but tricky problem—a Subsequence + DP variant wrapped in an "RPG game combo system".

Key points:

• Looks like simple "Longest Increasing Subsequence (LIS)".
• If you reflexively write an (O(N^2)) double loop, you're likely out.
• Only by utilizing the key constraint "fixed difference of 1" can you write the O(N) optimal solution expected by the interviewer.

With oavoservice's assistance, the candidate successfully jumped from a naive approach to the optimal solution and provided a space-friendly "path reconstruction" scheme in the follow-up, securing the next round for Google SDE Intern.

💬 Problem Requirements (Scenario Adapted)

Scenario: You are playing an RPG game. Each level has a difficulty value, given in an array order. Goal: You want to achieve the longest "Combo Chain". Rule: For every level you pick, the next level's difficulty must be exactly 1 greater than the previous one. You can skip levels, but must maintain original order. Task: Return the maximum length of such a Combo Chain.

Example

Input:  [10, 5, 11, 6, 7, 12, 8]
Output: 4
Explanation: Longest valid chain: 5 -> 6 -> 7 -> 8
# 10 -> 11 -> 12 has length 3, shorter.

🧠 Thought Process: From Clarification to State Definition

In the first 2 minutes, oavoservice guided the candidate through key Clarifications:

• Return 0 for empty input.
• Must indices be contiguous? No, skipping is allowed (subsequence, not subarray), but relative order must be preserved.
• Difficulty must be strictly +1? Yes, this constraint is crucial.

Many candidates would apply the LIS template here:

• For each nums[i], look back 0..i-1 for nums[j] == nums[i] - 1.
• dp[i] = max length ending at i.

Complexity: (O(N^2)). This might pass LeetCode, but in a Google interview, it signals "not sharp enough".

oavoservice's Key Hint: Compress to O(N) using "Fixed Difference 1"

We prompted a higher-level abstraction on the shared screen:

"This is a subsequence problem with fixed difference 1. Use a Hash Map to track 'longest chain ending at value'."

State definition:

• dp[x] = Length of the longest Combo Chain ending with difficulty x.

When traversing to difficulty x:

1. Check if x - 1 exists in our map.
2. If yes, extend that chain: dp[x] = dp[x - 1] + 1.
3. If no, start a new chain: dp[x] = 1.

One pass + Hash Map achieves O(N).

💻 Code Skeleton (Python)

def longest_combo(levels):
    dp = {}          # dp[x] = max length ending at val x
    max_combo = 0

    for level in levels:
        prev_length = dp.get(level - 1, 0)
        dp[level] = prev_length + 1
        max_combo = max(max_combo, dp[level])

    return max_combo

• Time Complexity: (O(N))
• Space Complexity: (O(N))

With oavoservice's hint, the candidate got it right immediately, skipping the N^2 detour. The interviewer nodded in approval.

🔁 Follow-up: How to Return the Specific Sequence?

The interviewer asked:

"If we need to return the specific sequence (e.g., [5, 6, 7, 8]) instead of just the length, what do we do?"

Common reaction:

• Store the whole list in dp.
• Pick the longest list at the end.

Result:

• Time (O(N)), but
• Space explodes to (O(N^2)). Not elegant.

oavoservice's Minimalist Reconstruction Idea

Our hint:

"Don't store the list. Just remember the end value of the max chain. The rest can be deduced."

Approach:

1. When updating max_combo, record the corresponding end_val.

   if dp[level] > max_combo:
       max_combo = dp[level]
       end_val = level
   

2. Knowing:

   • Max length L = max_combo
   • Ending difficulty end_val

   The chain must be:

   [end_val - L + 1, ..., end_val - 1, end_val]
   

3. Generate and reverse.

Candidate explanation:

"Since the step is strictly +1, once I know the max length and the ending value, I can reconstruct the sequence in O(L) just by subtracting 1 repeatedly."

Interviewer's comment: "Simple and efficient."

🎓 Interview Takeaways

• For subsequence problems with "fixed difference", think Hash Map, not double loops.
• Follow-ups often test space and representation. Don't blindly store paths; think if you can reverse-engineer from the end state.
• Interviews aren't about memorizing; they are about showing the evolution from naive to optimal solution.

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