0096EasyArrayHash TableStackMonotonic Stack

Auction Watchlist Highs

Tracked in this browser only

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.

An auction house sold its lots one after another and recorded the hammer price of each lot in sales, in the order the lots went under the hammer. No two lots fetched the same price.

A dealer kept a watchlist of prices, each of which is one of the recorded prices. For every price on the watchlist, the dealer wants to know the price of the first lot sold after that lot which fetched a strictly higher price. When no later lot beat it, the answer for that price is -1.

Return the answers in the same order as watchlist.

Examples

Example 1

Input: watchlist = [9, 3], sales = [3, 9, 5, 12]
Output: [12, 9]

The lot that fetched 9 was followed by 5, which is lower, then by 12, which beats it. The lot that fetched 3 was beaten right away by 9.

Example 2

Input: watchlist = [2, 10, 5], sales = [10, 2, 5, 7]
Output: [5, -1, 7]

After the lot at 2 came 5, which is higher. Nothing after the opening lot at 10 beat it, so its answer is -1. After 5 came 7.

Example 3

Input: watchlist = [6], sales = [9, 6, 1]
Output: [-1]

The lot at 9 sold before the watched lot, so it does not count; the only later lot fetched 1.

Constraints

1 <= watchlist.length <= sales.length <= 1000
0 <= watchlist[i], sales[i] <= 10^4
All prices within watchlist are distinct, and all prices within sales are distinct
Every price in watchlist appears in sales

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 next_higher_sale(watchlist: list[int], sales: list[int]) -> list[int]:

Java

public int[] nextHigherSale(int[] watchlist, int[] sales)

The intended approach

Time O(n + m)Space O(n)

Sweep `sales` once holding a stack of prices that are still waiting to be beaten, kept decreasing from bottom to top. Each new price resolves every waiting price below it, so pop and record the answer for those, then push the new price. Prices left on the stack at the end were never beaten. A dictionary from price to answer then serves the watchlist in one pass, which is sound because prices are distinct.

Pythonentry point: next_higher_sale

def next_higher_sale(watchlist, sales):
    answer_of = {}
    pending = []
    for price in sales:
        while pending and pending[-1] < price:
            answer_of[pending.pop()] = price
        pending.append(price)
    return [answer_of.get(price, -1) for price in watchlist]

The obvious version

For each watched price, locate its position in the sales log and walk forward until a strictly higher price turns up, O(n * m).

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_next_higher_sale(watchlist, sales):
    out = []
    for price in watchlist:
        start = sales.index(price)
        found = -1
        for j in range(start + 1, len(sales)):
            if sales[j] > price:
                found = sales[j]
                break
        out.append(found)
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in