1082HardArrayGreedySortingHeap (Priority Queue)

Rating a Group of Turbines

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Trains the technique from

LeetCode 1383Maximum Performance of a Team

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 wind farm owns n turbines. Turbine i produces output[i] units of power and runs at a steadiness of steadiness[i].

A group is any non-empty selection of at most k turbines. Its rating is the total output of the turbines in it multiplied by the smallest steadiness among them.

Return the largest rating a group can reach. Because that number can be large, return it modulo 10^9 + 7.

Examples

Example 1

Input: n = 3, output = [2, 4, 6], steadiness = [8, 5, 3], k = 2
Output: 30

The first two turbines total 6 in output with a smallest steadiness of 5, a rating of 30. The last two total 10 with a smallest steadiness of 3, also 30. Nothing does better.

Example 2

Input: n = 2, output = [10, 1], steadiness = [10, 1], k = 2
Output: 100

Adding the second turbine lifts the output to 11 but drags the smallest steadiness down to 1, rating 11. The first turbine on its own rates 100, so the best group is smaller than the limit allows.

Example 3

Input: n = 2, output = [10, 1], steadiness = [1, 100], k = 1
Output: 100

Only one turbine may be taken. The first rates 10 and the second rates 100.

Constraints

1 <= k <= n <= 10^5
output.length == n
steadiness.length == n
1 <= output[i] <= 10^5
1 <= steadiness[i] <= 10^8

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_performance(n: int, output: list[int], steadiness: list[int], k: int) -> int:

Java

public int maxPerformance(int n, int[] output, int[] steadiness, int k)

The intended approach

Time O(n log n) for the sort plus O(n log k) for the heap.Space O(n) for the order and O(k) for the heap.

A group's rating is decided by two very different things: a sum, which only grows as turbines join, and a minimum, which only shrinks. Fix the second and the first becomes easy. So walk the turbines in order of falling steadiness and, at each turbine, ask for the best group whose weakest steadiness is the one in hand. Every turbine already walked past is allowed in that group, and the multiplier is settled, so the job is simply to hold the largest total output over at most `k` of them. Keep the chosen outputs in a min-heap alongside their running sum. Push the current turbine's output, add it to the sum, and if the heap now holds more than `k` entries drop the smallest and take it off the sum. The heap therefore always holds the `k` biggest outputs seen so far, and multiplying its sum by the current steadiness gives one candidate rating. The answer is the largest candidate. A candidate is never an overestimate: if the current turbine happens to be the one just dropped, the sum belongs to turbines whose steadiness is at least the current one, so the real rating of that group is at least as big and was counted on its own turn. Reduce only at the very end, since the modulo would ruin the comparisons.

Pythonentry point: max_performance

def max_performance(n, output, steadiness, k):
    import heapq
    mod = 1000000007
    order = sorted(range(n), key=lambda i: -steadiness[i])
    heap = []
    total = 0
    best = 0
    for i in order:
        heapq.heappush(heap, output[i])
        total += output[i]
        if len(heap) > k:
            total -= heapq.heappop(heap)
        rating = total * steadiness[i]
        if rating > best:
            best = rating
    return best % mod

The obvious version

Try every group outright. Run through all selections of turbines as bit patterns, skip the ones holding more than `k` turbines, add up their output and track their smallest steadiness, and keep the largest rating. Costs two to the power of the turbine count, which only survives tiny farms.

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_performance(n, output, steadiness, k):
    mod = 1000000007
    best = 0
    for mask in range(1, 1 << n):
        size = 0
        total = 0
        low = None
        rest = mask
        i = 0
        while rest:
            if rest & 1:
                size += 1
                total += output[i]
                if low is None or steadiness[i] < low:
                    low = steadiness[i]
            rest >>= 1
            i += 1
        if size <= k:
            rating = total * low
            if rating > best:
                best = rating
    return best % mod

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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