1041HardArrayMathBinary SearchGreedySorting

How Many Growing Batches Can Be Filled

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Trains the technique from

LeetCode 2790Maximum Number of Groups With Increasing Length

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 store holds parts of several kinds, and stock[i] is how many parts of kind i it holds.

Fill a run of batches so that no batch holds two parts of the same kind, no kind is drawn on more times across all the batches than the store holds, and each batch after the first holds strictly more parts than the one before it.

Return the greatest number of batches that can be filled.

Examples

Example 1

Input: stock = [5, 2, 1]
Output: 3

Three batches need one, two and three parts, six in all. The batch of three takes one part of each kind, the batch of two takes one of the first kind and one of the second, and the batch of one takes another of the first kind. A fourth batch would need four different kinds and only three exist.

Example 2

Input: stock = [3, 3]
Output: 2

Only two kinds exist, so no batch can hold three parts without repeating a kind, which caps the run at a batch of one and a batch of two.

Example 3

Input: stock = [1000000000]
Output: 1

One kind means no batch can hold two parts, so a single batch of one part is all there is, however many parts the store holds.

Constraints

1 <= stock.length <= 10^5
1 <= stock[i] <= 10^9

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_increasing_groups(stock: list[int]) -> int:

Java

public int maxIncreasingGroups(int[] stock)

The intended approach

Time O(n log n) for the sort, then one pass.Space O(n) for the sorted stocks.

Suppose `k` batches are filled. Their sizes climb strictly, so the smallest they can be altogether is one, two, and so on up to `k`, which adds to half of `k` times `k` plus one. Call that the triangle for `k`. The store's total has to reach it, so the triangle is a hard floor. Sort the stocks upward and walk them, carrying the running total. Each time the running total reaches the triangle for one more batch than currently claimed, claim one more. What stands claimed at the end is the answer. Two things make that work. The triangle floor means nothing can be claimed too early. And with the stocks sorted upward, once the total does reach the triangle for `k`, the parts counted so far really can be laid out: go through the sorted kinds and spread each kind's parts one to a batch, wrapping round the batches, so a kind with more parts than there are batches is simply capped at one part per batch. Sorting upward puts those large kinds last, once the batch count has grown as far as it will, so nothing a kind offers is wasted. Note also that the count is claimed at most once per kind, which is what keeps a batch from being asked to hold more distinct kinds than exist. So a sort and one pass.

Pythonentry point: max_increasing_groups

def max_increasing_groups(stock):
    order = sorted(stock)
    total = 0
    batches = 0
    for count in order:
        total += count
        if total >= (batches + 1) * (batches + 2) // 2:
            batches += 1
    return batches

The obvious version

Test each batch count outright, climbing from one, and keep the largest that passes. Batches of sizes one up to `k` can be filled exactly when no group of batches ever demands more than the kinds can supply, and the demanding groups are the largest batches: for every count of batches from one to `k`, the sizes of that many largest batches added together must not exceed what the kinds offer when each kind is capped at one part per batch in the group. Checking every such group is a different test altogether from the running triangle.

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_increasing_groups(stock):
    n = len(stock)
    def feasible(k):
        if k > n:
            return False
        for s in range(1, k + 1):
            demand = s * k - s * (s - 1) // 2
            room = 0
            for count in stock:
                room += count if count < s else s
            if demand > room:
                return False
        return True
    best = 0
    k = 1
    while feasible(k):
        best = k
        k += 1
    return best

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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