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Assessment reportAccepted

Problem: Max score in k movesMedium · Dynamic programming
Correctness: 14 of 14 tests passedIncluding 4 hidden edge cases
Complexity: O(n·k) time, O(k) spaceMeets the target bound

Reviewer notes

Strength: Clean state: dp[j] is the best total with j moves left.
Improve: Say why a 1-D table is enough before you write it.
Follow-up: Moves may also step back one index. What changes?

Recommended nextSliding window maximumHard

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MediumDynamic programmingJane Street · trading round

Max score in k moves

You start at index 0 of an integer array arr. On each move you may stay or advance to i + 1. Before every move you collect arr[i]. Return the maximum total achievable after exactly k moves.

Example

arr = [1, 20, 3], k = 3  // -> 41
// collect 1, advance; collect 20, stay; collect 20

Constraints

1 ≤ n, k ≤ 10⁵
arr[i] may be negative

Try a problem free

EditorResultsExplanationC++23 · GCC 14

long long maxScore(vector<int>& arr, int k) {
    int n = arr.size();
    // dp[j] = best total using exactly j moves from here
    vector<long long> dp(k + 1, 0);
    for (int i = n - 1; i >= 0; --i) {
        long long best = 0;
        for (int j = 1; j <= k; ++j) {
            // collect arr[i], then stay or advance
            long long stay = arr[i] + dp[j - 1];
            best = max(best, stay);
            dp[j] = best;
        }
    }
    return dp[k];
}

#1 arr=[1,2,3] k=30.3ms · Accepted

#2 arr=[1,20,3] k=30.5ms · Accepted

#3 arr=[5,-2,9] k=40.4ms · Accepted

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Mock interviewTwo Sum · follow-ups12:48

Interviewer

Your hash-map solution is O(n) time. What changes if the numbers arrive as a stream you can’t store?

You

I’d keep a bounded window of recent values, or trade exactness for memory with a Bloom filter.

Interviewer

Good. How does the false-positive rate behave as the filter fills up?

Listening to your answer…

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Edge cases

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