0963EasyArrayBinary SearchSegment TreeSimulationOrdered Set

Crates That Find No Bin

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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.

Crate sizes are given as fruits and bin capacities as baskets, both of the same length.

The crates are dealt with in the order listed. Each crate goes into the leftmost bin not yet used whose capacity is at least the crate's size; a bin holds at most one crate. A crate with no such bin left is unplaced.

Return how many crates end up unplaced.

Examples

Example 1

Input: fruits = [14, 3, 27, 9], baskets = [10, 30, 5, 20]
Output: 1

The crate of 14 takes the bin of 30, since the bin of 10 is too small. The crate of 3 takes the bin of 10. The crate of 27 finds nothing big enough left, and the crate of 9 takes the bin of 20.

Example 2

Input: fruits = [3, 1], baskets = [1, 3]
Output: 0

The crate of 3 takes the second bin, and the crate of 1 takes the first, which was passed over as too small the first time.

Example 3

Input: fruits = [7, 7, 7], baskets = [6, 6, 6]
Output: 3

Every bin is smaller than every crate, so nothing can be placed.

Constraints

fruits.length == baskets.length
1 <= fruits.length <= 100
1 <= fruits[i] <= 1000
1 <= baskets[i] <= 1000

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^2) for n crates: each one may scan every bin.Space O(n) for the used marks.

The rule fixes everything: crates in the order given, and for each one the leftmost bin still free that is big enough. There is no choice to make and nothing to optimise, so the answer follows from walking it through. Keep a used mark per bin, and for each crate scan the bins from the left until a free one large enough turns up, marking it used; a scan that finds nothing means the crate is unplaced. With a hundred crates at most, scanning the bins per crate is affordable. Note that a bin passed over as too small for one crate may well take a later, smaller one, so bins are never retired.

Pythonentry point: num_of_unplaced_fruits

def num_of_unplaced_fruits(fruits, baskets):
    n = len(fruits)
    used = [False] * n
    unplaced = 0
    for size in fruits:
        placed = False
        for j in range(n):
            if used[j]:
                continue
            if baskets[j] >= size:
                used[j] = True
                placed = True
                break
        if not placed:
            unplaced += 1
    return unplaced

The obvious version

For each crate, build the list of bins still free from scratch, walk it in order looking for the first big enough, and take that bin out of the pool. Rebuilds the pool per crate rather than keeping marks.

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):
    free = []
    for j in range(len(baskets)):
        free.append([j, baskets[j]])
    unplaced = 0
    for size in fruits:
        pool = sorted(free)
        pick = None
        for entry in pool:
            if entry[1] >= size:
                pick = entry
                break
        if pick is None:
            unplaced += 1
        else:
            free.remove(pick)
    return unplaced

How this problem was checked

Reference solution executed against 10 test cases — 10 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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