0535MediumArrayBinary SearchGreedySliding WindowSortingPrefix Sum

Richest Run of Consecutive Bins

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Trains the technique from

LeetCode 3413Maximum Coins From K Consecutive Bags

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 picking aisle has bins at every whole position 1, 2, 3, ... running away from the door. The stock is recorded in stretches: runs[i] = [from, to, per] means every bin at a position from from to to inclusive holds exactly per items. No two stretches cover the same position, and a position no stretch mentions holds nothing.

A picker is sent down the aisle to clear span bins standing next to each other, that is span consecutive positions, all of them at position 1 or beyond. Return the largest number of items such a group of bins can hold.

Examples

Example 1

Input: runs = [[2, 5, 3], [8, 9, 10]], span = 4
Output: 20

Clearing positions 6 through 9 picks up nothing at 6 and 7 and 10 items at each of 8 and 9, for 20 items.

Example 2

Input: runs = [[1, 10, 1], [11, 12, 100]], span = 3
Output: 201

Clearing positions 10, 11 and 12 picks up 1 item at position 10 and 100 at each of 11 and 12, for 201 items.

Example 3

Input: runs = [[5, 5, 7]], span = 3
Output: 7

Only position 5 holds anything, so any group of three bins covering it, such as positions 4, 5 and 6, picks up 7 items.

Constraints

1 <= runs.length <= 10^5
1 <= span <= 10^9
runs[i].length == 3
1 <= runs[i][0] <= runs[i][1] <= 10^9
1 <= runs[i][2] <= 1000
The stretches do not overlap.
The answer is at most span * 1000, so it stays below 10^12.

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 richest_run(runs: list[list[int]], span: int) -> int:

Java

public long richestRun(int[][] runs, int span)

The intended approach

Time O(n log n) for n stretches: one sort, then a constant number of binary searches per stretch.Space O(n) for the sorted stretches and their running totals.

Sort the stretches by their first position and build a running total of the items they hold, so the items inside any group of stretches is one subtraction. A window then costs two binary searches, one for the first stretch reaching into it and one for the last stretch starting inside it, plus a correction for the part of each of those two stretches that sticks out. Sliding a window sideways only changes what it holds when an end passes a stretch boundary, so it is enough to test the windows whose left edge sits on a stretch's first position and the windows whose right edge sits on a stretch's last position.

Pythonentry point: richest_run

def richest_run(runs, span):
    # O(n log n): sort the runs, prefix-total them, and test only the windows
    # that begin at a run's first bin or finish at a run's last bin.
    from bisect import bisect_left, bisect_right
    segs = sorted(runs, key=lambda seg: seg[0])
    starts = [seg[0] for seg in segs]
    ends = [seg[1] for seg in segs]
    pre = [0] * (len(segs) + 1)
    for i in range(len(segs)):
        pre[i + 1] = pre[i] + (segs[i][1] - segs[i][0] + 1) * segs[i][2]

    def window(left):
        right = left + span - 1
        i = bisect_left(ends, left)
        j = bisect_right(starts, right) - 1
        if i > j:
            return 0
        total = pre[j + 1] - pre[i]
        if segs[i][0] < left:
            total -= (left - segs[i][0]) * segs[i][2]
        if segs[j][1] > right:
            total -= (segs[j][1] - right) * segs[j][2]
        return total

    best = 0
    for seg in segs:
        got = window(seg[0])
        if got > best:
            best = got
        got = window(seg[1] - span + 1)
        if got > best:
            best = got
    return best

The obvious version

Write out how much every mentioned position holds into a lookup, then try every possible left edge from well before the first stretch to just past the last one, adding up the span positions one at a time.

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_richest_run(runs, span):
    holds = {}
    for seg in runs:
        for pos in range(seg[0], seg[1] + 1):
            holds[pos] = seg[2]
    lo = min(seg[0] for seg in runs)
    hi = max(seg[1] for seg in runs)
    best = 0
    for left in range(lo - span, hi + 2):
        if left < 1:
            continue
        total = 0
        for pos in range(left, left + span):
            total += holds.get(pos, 0)
        if total > best:
            best = total
    return best

How this problem was checked

Reference solution executed against 12 test cases — 12 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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