1191EasyArrayHash TableSliding WindowHeap (Priority Queue)

The Loudest Tracks in Every Stretch

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Trains the technique from

LeetCode 3318Find X-Sum of All K-Long Subarrays I

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 jukebox log lists what was played, plays[i] being the track number of the i-th play.

The weight of a stretch of plays is worked out like this: tally how often each track turns up in the stretch, then keep the keep tracks with the highest tallies, settling a tie in favour of the higher track number. The weight is the total of each kept track's number times its tally in the stretch. When the stretch holds keep or fewer distinct tracks they are all kept, so its weight is the total of the stretch.

Return an array whose i-th entry is the weight of the stretch of span plays starting at position i.

Examples

Example 1

Input: plays = [4, 4, 5, 6, 6, 6], span = 3, keep = 1
Output: [8, 6, 12, 18]

Four stretches of three plays. The first tallies track 4 twice against track 5 once and keeps track 4 for 8. The second tallies all three tracks once, so the tie hands it to track 6. The third keeps track 6 with two plays for 12, and the last is three plays of track 6.

Example 2

Input: plays = [2, 2, 3, 3], span = 4, keep = 1
Output: [6]

One stretch, in which tracks 2 and 3 are both tallied twice. The tie goes to the higher track number, so track 3 is kept for two plays.

Example 3

Input: plays = [1, 1, 1, 9], span = 4, keep = 2
Output: [12]

The stretch holds only two distinct tracks and both are kept, so the weight is the total of the whole stretch.

Constraints

1 <= plays.length <= 50
1 <= plays[i] <= 50
1 <= keep <= 50
1 <= span <= 50
keep <= span <= plays.length

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 find_x_sum(plays: list[int], span: int, keep: int) -> list[int]:

Java

public int[] findXSum(int[] plays, int span, int keep)

The intended approach

Time O(n * span * log(span)) over the stretches.Space O(span) for one window's tallies.

The log is short enough to take each stretch on its own terms. Slide a window of `span` plays along it and, for each position, tally the tracks inside the window with a hash map. Ordering the tallied tracks decides what is kept. Sort them by falling tally and, where two tallies match, by falling track number, which is exactly the tie rule. Then take the first `keep` of that order and add up track number times tally over them. Taking the first `keep` also covers the case of few distinct tracks on its own: when the window holds no more than `keep` of them, the whole order is taken and the weight comes out as the plain total of the window, so no separate branch is needed.

Pythonentry point: find_x_sum

def find_x_sum(plays, span, keep):
    weights = []
    for start in range(len(plays) - span + 1):
        tally = {}
        for at in range(start, start + span):
            track = plays[at]
            tally[track] = tally.get(track, 0) + 1
        order = sorted(tally.items(), key=lambda pair: (-pair[1], -pair[0]))
        total = 0
        for track, count in order[:keep]:
            total += track * count
        weights.append(total)
    return weights

The obvious version

Pick the kept tracks out one at a time. For a stretch, count each distinct track by scanning the window for it, then repeat `keep` times: sweep the surviving tracks for the one with the largest tally, taking the higher track number when tallies match, add its number times its tally to the running weight and drop it from the survivors. Stop early when nothing is left. No sorting anywhere, just repeated sweeps for the best remaining track.

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_find_x_sum(plays, span, keep):
    weights = []
    for start in range(len(plays) - span + 1):
        window = plays[start:start + span]
        kinds = []
        for track in set(window):
            count = 0
            for play in window:
                if play == track:
                    count += 1
            kinds.append([track, count])
        total = 0
        taken = 0
        while taken < keep and kinds:
            best = 0
            for i in range(1, len(kinds)):
                if kinds[i][1] > kinds[best][1]:
                    best = i
                elif kinds[i][1] == kinds[best][1] and kinds[i][0] > kinds[best][0]:
                    best = i
            track, count = kinds.pop(best)
            total += track * count
            taken += 1
        weights.append(total)
    return weights

How this problem was checked

Reference solution executed against 15 test cases — 15 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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