Sort Array by Increasing Frequency
Real-world version: Order Readings by Repeat CountThe statement here is our own: the same inputs, outputs and technique.
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 starts with this. It matches the parent problem's shape, so a solution that works here works on a standard judge unchanged.
def frequency_sort(nums: list[int]) -> list[int]:public int[] frequencySort(int[] nums)Checking your plan…
How this problem was checked
- Passed: Reference solution run against 9 test cases: 9 passed
- Passed: 3 worked examples confirmed by running the code
- Passed: 0 eight-word sequences shared with the parent problem's text
- Passed: Agrees with an independent brute-force solution on random inputs
- Passed: Test set shown to catch deliberately broken solutions
- Passed: Reviewed by a separate quality pass
- Passed: Function arity matches the parent problem's