0578EasyArrayHash TableSorting

Order Readings by Repeat Count

Tracked in this browser only

Trains the technique from

LeetCode 1636Sort Array by Increasing Frequency

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 calibration rig logs temperature drifts into nums, where a drift may be negative, zero or positive and the same drift may be logged many times.

Rearrange the whole log so that a drift logged fewer times stands ahead of a drift logged more times. When two drifts were logged the same number of times, the larger drift comes first. Each drift must still appear exactly as many times as it did in the input, and all copies of one drift sit together.

Return the rearranged log.

Examples

Example 1

Input: nums = [4, -2, 4, -2, 9]
Output: [9, 4, 4, -2, -2]

9 was logged once, 4 twice and -2 twice. The single 9 leads, then the two drifts tied on two logs each appear larger first, so both 4s precede both -2s.

Example 2

Input: nums = [-100, 100, -100, 100, -100, 3]
Output: [3, 100, 100, -100, -100, -100]

3 was logged once, 100 twice and -100 three times, so the groups come out in that order of tally.

Example 3

Input: nums = [6, 6, 2, 2, 9, 9]
Output: [9, 9, 6, 6, 2, 2]

All three drifts were logged twice, so the tie rule alone decides and they come out from largest to smallest.

Constraints

1 <= nums.length <= 100
-100 <= 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 frequency_sort(nums: list[int]) -> list[int]:

Java

public int[] frequencySort(int[] nums)

The intended approach

Time O(n log n) — one tally pass plus one comparison sort of the n entries.Space O(n) for the tally and the returned log.

Walk the log once building a dictionary from drift to how often it was logged. Then sort the log itself, ranking each entry by the pair (its tally, its negated value): the tally puts rare drifts first and the negation turns the secondary comparison into a decreasing one, so equal tallies come out largest first. Sorting the entries rather than the distinct values means the copies land grouped automatically.

Pythonentry point: frequency_sort

def frequency_sort(nums):
    # O(n log n) time, O(n) space: one pass tallies how often each reading
    # occurs, then a single sort uses the tally as the first key and the
    # negated reading as the second so equal tallies fall in decreasing order.
    count = {}
    for v in nums:
        count[v] = count.get(v, 0) + 1
    return sorted(nums, key=lambda v: (count[v], -v))

The obvious version

Collect the distinct drifts and their tallies, then place them one at a time into a list by insertion, scanning forward past every drift that should precede the new one. Finally expand each drift into as many copies as its 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_frequency_sort(nums):
    counts = {}
    distinct = []
    for v in nums:
        if v not in counts:
            distinct.append(v)
            counts[v] = 0
        counts[v] += 1
    order = []
    for v in distinct:
        pos = 0
        while pos < len(order):
            u = order[pos]
            if counts[u] < counts[v] or (counts[u] == counts[v] and u > v):
                pos += 1
            else:
                break
        order.insert(pos, v)
    out = []
    for v in order:
        for _ in range(counts[v]):
            out.append(v)
    return out

How this problem was checked

Reference solution executed against 9 test cases — 9 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in