0633EasyArraySortingHeap (Priority Queue)

Best Pair Of Voucher Payouts

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Trains the technique from

LeetCode 1464Maximum Product of Two Elements in an Array

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 shop prints a run of vouchers. Voucher i shows the face value tokens[i], and handing it in converts to tokens[i] - 1 credits. Two vouchers may carry the same face value.

Choose two vouchers at different positions i and j. The payout for that choice is the product of their credit amounts. Return the largest payout any such choice gives.

Examples

Example 1

Input: tokens = [9, 4, 7, 1, 10]
Output: 72

Taking the vouchers showing 10 and 9 gives credits of 9 and 8, and their product is 72.

Example 2

Input: tokens = [6, 6, 2]
Output: 25

The two vouchers showing 6 sit at different positions, so both may be used. Each converts to 5 credits, and 5 times 5 is 25.

Example 3

Input: tokens = [1, 1000]
Output: 0

Only one pair exists. The voucher showing 1 converts to 0 credits, so the payout is 0 times 999, which is 0.

Constraints

2 <= tokens.length <= 500
1 <= tokens[i] <= 10^3

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_product(tokens: list[int]) -> int:

Java

public int maxProduct(int[] tokens)

The intended approach

Time O(n) — a single pass with constant work per voucher.Space O(1) — two running values.

Face values are at least 1, so every credit amount is at least 0 and the payout never falls when a factor grows. One pass keeps the largest face value and the largest face value at a different position, updating both as values arrive, and the answer is the product of the two credit amounts. Equal face values are kept separately rather than folded together, so a run printed twice at the same value still fills both slots.

Pythonentry point: max_product

def max_product(tokens):
    # O(n) time, O(1) space: the payout rises with either factor, so one pass
    # keeping the largest and the second largest face value is enough. Positions
    # are tracked by value, not by identity, so two vouchers printed with the
    # same face value still supply both factors.
    top = 0
    second = 0
    for v in tokens:
        if v > top:
            second = top
            top = v
        elif v > second:
            second = v
    return (top - 1) * (second - 1)

The obvious version

Loop over every unordered pair of positions, work out the payout for that pair directly, and keep the largest seen.

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_product(tokens):
    best = None
    for i in range(len(tokens)):
        for j in range(i + 1, len(tokens)):
            pay = (tokens[i] - 1) * (tokens[j] - 1)
            if best is None or pay > best:
                best = pay
    return best

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

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