0635MediumArrayHash TableSliding Window

Richest Run Of Distinct Bins

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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 picking aisle is a row of bins. bins[i] is the part number stored in bin i, and clearing a bin earns that many points. The same part number may be stored in more than one bin.

A picker clears one contiguous stretch of bins, and the audit only accepts the run when no part number appears twice inside it. Return the largest number of points such a run can earn.

Examples

Example 1

Input: bins = [1, 5, 4, 5]
Output: 10

Clearing bins 0 through 2 collects part numbers 1, 5 and 4, which are all different, for 10 points. Adding bin 3 would repeat part number 5, so that run is not accepted.

Example 2

Input: bins = [4, 2, 7, 9]
Output: 22

No part number appears twice anywhere, so the whole aisle is one accepted run worth 4 + 2 + 7 + 9 = 22 points.

Example 3

Input: bins = [2, 1, 2, 1, 2, 1, 2]
Output: 3

Any accepted run holds at most one bin of part number 2 and one of part number 1, so a run such as bins 0 and 1 earns 3 points.

Constraints

1 <= bins.length <= 10^5
1 <= bins[i] <= 10^4

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 maximum_unique_subarray(bins: list[int]) -> int:

Java

public int maximumUniqueSubarray(int[] bins)

The intended approach

Time O(n) — each bin joins the window once and leaves it at most once.Space O(n) for the set of part numbers inside the window.

Slide a window along the row that always holds distinct part numbers, carrying both a membership set and the sum of the values inside. When the next bin repeats something already in the window, keep dropping bins off the left end, one at a time and subtracting each from the sum, until the repeat is gone; several bins may have to go on a single step. The bin then joins the window and the running sum is compared with the best seen.

Pythonentry point: maximum_unique_subarray

def maximum_unique_subarray(bins):
    # O(n) time, O(n) space: hold a window over the row that never repeats a part
    # number, together with its running total. Before the bin at hi joins, drop
    # bins off the left end until its part number is no longer inside, taking
    # each dropped value off the total, then admit hi and note the best total.
    inside = set()
    total = 0
    best = 0
    lo = 0
    for hi in range(len(bins)):
        v = bins[hi]
        while v in inside:
            total -= bins[lo]
            inside.discard(bins[lo])
            lo += 1
        inside.add(v)
        total += v
        if total > best:
            best = total
    return best

The obvious version

Start a fresh run at every bin, walk right while the part numbers stay distinct, add up as you go, and stop that run at the first repeat. Keep the largest total any run reaches.

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_maximum_unique_subarray(bins):
    n = len(bins)
    best = 0
    for start in range(n):
        seen = set()
        total = 0
        for end in range(start, n):
            if bins[end] in seen:
                break
            seen.add(bins[end])
            total += bins[end]
            if total > best:
                best = total
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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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