0774MediumArrayBinary SearchSegment TreeOrdered Set

Crates Into Rated Shelf Slots

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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 depot has a row of shelf slots, left to right, and baskets[j] is the load rating of slot j in kilograms. Crates arrive one at a time in the order given by fruits, and fruits[i] is the weight of the i-th crate.

Each arriving crate is handled by the same rule: scan the slots from left to right and put the crate in the first still-empty slot whose rating is greater than or equal to the crate's weight. A slot holds at most one crate, and once a crate is in a slot it is never moved. If no still-empty slot is rated high enough, that crate is turned away and the slots are left as they were before it arrived.

Return how many crates end up turned away.

Examples

Example 1

Input: fruits = [4,2,5], baskets = [3,5,4]
Output: 1

The 4 kg crate skips slot 0, rated 3, and lands in slot 1, rated 5. The 2 kg crate then takes slot 0, which is still empty and rated high enough. The 5 kg crate finds only slot 2 free, rated 4, so it is turned away.

Example 2

Input: fruits = [6,6,1], baskets = [6,2,9]
Output: 0

The first 6 kg crate takes slot 0, whose rating of 6 exactly matches it. The second 6 kg crate passes slot 1, rated 2, and takes slot 2. The 1 kg crate takes slot 1, so no crate is turned away.

Example 3

Input: fruits = [2,2,2], baskets = [9,1,1]
Output: 2

The first 2 kg crate takes slot 0. Slots 1 and 2 are both rated 1, below 2 kg, so the other two crates are turned away.

Constraints

n == fruits.length
n == baskets.length
fruits.length == baskets.length
1 <= fruits.length <= 10^5
1 <= baskets.length <= 10^5
1 <= fruits[i] <= 10^9
1 <= baskets[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 num_of_unplaced_fruits(fruits: list[int], baskets: list[int]) -> int:

Java

public int numOfUnplacedFruits(int[] fruits, int[] baskets)

The intended approach

Time O(n log n) for n crates and n slots: one root-to-leaf descent plus one upward repair per crate.Space O(n) for the segment tree.

Hold the slot ratings in a segment tree whose every internal node keeps the maximum rating in its range. For a crate, the root maximum says at once whether any free slot can take it; if one can, walk down from the root always stepping into the left child when that child's maximum is at least the crate weight and into the right child otherwise, which lands on the leftmost qualifying slot. Set that leaf to `0` and repair the maxima up to the root, so the slot can never be chosen again because crate weights start at `1`.

Pythonentry point: num_of_unplaced_fruits

def num_of_unplaced_fruits(fruits, baskets):
    # Segment tree over slot ratings holding the maximum of each range, so the
    # leftmost slot that can take a crate is found by one descent.
    # Time O(n log n), space O(n).
    n = len(baskets)
    width = 1
    while width < n:
        width *= 2
    tree = [0] * (2 * width)
    for i, cap in enumerate(baskets):
        tree[width + i] = cap
    for i in range(width - 1, 0, -1):
        tree[i] = tree[2 * i] if tree[2 * i] > tree[2 * i + 1] else tree[2 * i + 1]

    rejected = 0
    for load in fruits:
        if tree[1] < load:
            rejected += 1
            continue
        node = 1
        while node < width:
            node *= 2
            if tree[node] < load:
                node += 1
        # A taken slot is set to 0, which no crate can match since loads start at 1.
        tree[node] = 0
        node //= 2
        while node:
            tree[node] = tree[2 * node] if tree[2 * node] > tree[2 * node + 1] else tree[2 * node + 1]
            node //= 2
    return rejected

The obvious version

Keep a boolean per slot saying whether it is filled, and for each crate walk the slots from the left until one is both free and rated high enough. Quadratic in the number of crates but it applies the placement rule literally.

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_num_of_unplaced_fruits(fruits, baskets):
    taken = [False] * len(baskets)
    rejected = 0
    for load in fruits:
        placed = False
        for i, cap in enumerate(baskets):
            if not taken[i] and cap >= load:
                taken[i] = True
                placed = True
                break
        if not placed:
            rejected += 1
    return rejected

How this problem was checked

Reference solution executed against 14 test cases — 14 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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