0787HardArrayDynamic ProgrammingGreedySorting

Running Order For The Demo Reel

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A showcase reel is cut together from a pile of clips. Clip i has been given a rating ratings[i] by the review panel, and a rating may be negative when the panel disliked the clip.

You choose which clips go into the reel and what order they play in. Any number of clips may be left out, including all of them. A clip that ends up in position p of the finished reel scores p * rating, where positions are numbered 1, 2, 3, ... over the clips that were actually included, so leaving a clip out shifts everything after it one position earlier. The reel's value is the sum of the scores of its clips, and an empty reel is worth 0.

Return the largest value a reel cut from these clips can have.

Examples

Example 1

Input: ratings = [-4, 1, 2]
Output: 5

Leave the clip rated -4 out and play the other two as 1 then 2. That reel scores 1 * 1 + 2 * 2 = 5.

Example 2

Input: ratings = [-1, 5, 6]
Output: 27

Include all three clips in the order -1, 5, 6. The scores are 1 * (-1), 2 * 5 and 3 * 6, which add up to 27.

Example 3

Input: ratings = [-7, -2, -3]
Output: 0

Every clip is rated below zero. Leaving all of them out gives an empty reel, worth 0.

Constraints

n == ratings.length
1 <= n <= 500
-1000 <= ratings[i] <= 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def max_reel_value(ratings: list[int]) -> int:

Java

public int maxReelValue(int[] ratings)

The intended approach

Time O(n log n): one sort of n ratings, then a single pass over the sorted list.Space O(n) for the sorted copy of the ratings.

Sort the ratings ascending, then sweep the sorted list from its top end while keeping a running sum of the clips accepted so far. Accepting the next clip puts it in front of everything already accepted, which pushes each of those one position later and so gains the running sum a second time in addition to the new clip's own rating; the sweep stops the moment that combined figure is no longer positive. The answer is the sum of the running sums recorded along the way, which is 0 when the sweep stops at once.

Pythonentry point: max_reel_value

def max_reel_value(ratings):
    # Sort ascending and consider the clips from the highest rating downwards,
    # keeping a running sum of the clips taken so far. Adding one more clip in
    # front of the current selection shifts every kept clip one position later,
    # which adds the running sum again on top of the new clip's own rating; the
    # scan stops as soon as that combined figure stops being an improvement.
    # Time O(n log n) for the sort plus one pass, space O(n) for the sorted copy.
    order = sorted(ratings)
    total = 0
    run = 0
    for value in reversed(order):
        if run + value <= 0:
            break
        run += value
        total += run
    return total

The obvious version

Enumerate every subset of clips as a bit mask and every ordering of each subset, score each reel directly from the definition, and keep the largest score found alongside the empty reel's 0.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_max_reel_value(ratings):
    from itertools import permutations

    n = len(ratings)
    best = 0
    for mask in range(1 << n):
        chosen = [ratings[i] for i in range(n) if (mask >> i) & 1]
        for arrangement in permutations(chosen):
            score = 0
            for pos, value in enumerate(arrangement, start=1):
                score += pos * value
            if score > best:
                best = score
    return best

How this problem was checked

Reference solution executed against 16 test cases — 16 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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