0328EasyArrayHash TableSortingCounting Sort

Bake Off Scores Beaten

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Trains the technique from

LeetCode 1365How Many Numbers Are Smaller Than the Current Number

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 bake-off has finished and nums[i] is the panel score awarded to entry i, on a scale from 0 to 100.

The organisers want a card for every entry saying how many entries it beat outright. For entry i that is the number of positions j with j != i and nums[j] < nums[i]. Entries that tie on score beat each other zero times.

Return an array out of the same length as nums, where out[i] is the count for entry i, in the same order as the input.

Examples

Example 1

Input: nums = [7, 2, 7, 4]
Output: [2, 0, 2, 1]

Each 7 beats the 2 and the 4 but not the other 7, so both get 2; the 2 beats nothing; the 4 beats only the 2.

Example 2

Input: nums = [6, 6, 6, 6, 6]
Output: [0, 0, 0, 0, 0]

All five entries tie, so none of them beats another and every card reads 0.

Example 3

Input: nums = [0, 100]
Output: [0, 1]

The lowest score beats nothing and the highest beats the one entry below it.

Example 4

Input: nums = [9, 1, 5, 1, 9, 0]
Output: [4, 1, 3, 1, 4, 0]

Each 9 sits above the 1, 5, 1 and 0, giving 4; each 1 sits above only the 0; the 5 sits above the two 1s and the 0.

Constraints

2 <= nums.length <= 500
0 <= nums[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 strictly_below_counts(points: list[int]) -> list[int]:

Java

public int[] strictlyBelowCounts(int[] points)

The intended approach

Time O(n + V) where V = 101 possible scoresSpace O(V) for the tally plus O(n) for the returned array

The score range is fixed by the constraints, so one pass fills a tally of how many entries hold each score. A second pass over the score range turns that tally into, for each score, the number of entries strictly under it, and then each input score is looked up directly.

Pythonentry point: strictly_below_counts

def strictly_below_counts(points):
    # Scores live in a fixed window, so tally them and turn the tally into a
    # running count of everything strictly under each score.
    # Time O(n + V) with V = 101 score values. Space O(V) plus the O(n) answer.
    tally = [0] * 101
    for v in points:
        tally[v] += 1
    below = [0] * 101
    running = 0
    for v in range(101):
        below[v] = running
        running += tally[v]
    return [below[v] for v in points]

The obvious version

For every entry, walk the whole array again and count the lower scores.

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_strictly_below_counts(points):
    out = []
    for a in points:
        c = 0
        for b in points:
            if b < a:
                c += 1
        out.append(c)
    return out

How this problem was checked

Reference solution executed against 7 test cases — 7 passed
4 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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