1107EasyArraySortingCounting SortBubble Sort

Jars Standing in the Wrong Place

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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 shelf holds jars whose sizes are listed in sizes, left to right. The jars are meant to stand in non-decreasing order of size.

Return how many positions on the shelf hold a jar of a different size from the one that would stand there once the jars were put in that order.

Examples

Example 1

Input: sizes = [2, 1, 2, 1]
Output: 2

In size order the jars would stand 1, 1, 2, 2. Every one of the four positions holds a different size from that.

Example 2

Input: sizes = [1, 3, 2, 4, 6, 5]
Output: 4

In size order the jars would run 1 up to 6. The second and third positions are the wrong way round and so are the last two.

Example 3

Input: sizes = [1, 1, 1, 1]
Output: 0

Every jar is the same size, so the shelf already stands in size order.

Constraints

1 <= sizes.length <= 100
1 <= sizes[i] <= 100

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 height_checker(sizes: list[int]) -> int:

Java

public int heightChecker(int[] sizes)

The intended approach

Time O(n + range of sizes).Space O(range of sizes) for the tallies.

The question is not about neighbours at all, which is what makes counting places where a jar is bigger than the one to its left the wrong answer. It compares the shelf against one particular arrangement, so that arrangement has to be produced and walked alongside the shelf. The sizes are small whole numbers, so tally how many jars carry each size and then read the tallies upwards. That hands out the sizes in exactly the order they would stand in, one at a time, and each one can be compared against the jar actually standing at that position as it is handed out. Every mismatch counts, so a size appearing several times contributes a comparison for each copy. Counting the sizes rather than sorting the jars keeps the work to the shelf length plus the range of sizes.

Pythonentry point: height_checker

def height_checker(sizes):
    top = 0
    for value in sizes:
        if value > top:
            top = value
    tally = [0] * (top + 1)
    for value in sizes:
        tally[value] += 1
    wrong = 0
    at = 0
    for value in range(top + 1):
        for _ in range(tally[value]):
            if sizes[at] != value:
                wrong += 1
            at += 1
    return wrong

The obvious version

Take a copy of the sizes, sort it outright, then walk the shelf and the sorted copy side by side counting the positions where they disagree. Costs a sort rather than a tally.

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_height_checker(sizes):
    wanted = sorted(sizes)
    wrong = 0
    for i in range(len(sizes)):
        if sizes[i] != wanted[i]:
            wrong += 1
    return wrong

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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