0011MediumArrayHash TableCountingBucket SortHeap (Priority Queue)Sorting

Most Reported Calibration Offsets

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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 field probe emits one signed calibration offset each time it is polled. The array readings holds every offset captured during a single shift, and k is how many of them a technician wants to look at.

Report the k distinct offsets that were emitted the greatest number of times. Those k values may be handed back in whatever order is convenient.

The shift log is filtered before it reaches you so the choice is never ambiguous: any offset left out of the answer was emitted strictly fewer times than any offset placed in it.

Examples

Example 1

Input: readings = [4, -2, 4, 4, -2, 7], k = 2
Output: [4, -2]

Offset 4 lands three times and offset -2 twice, both ahead of 7 with its single appearance.

Example 2

Input: readings = [3, 3, 8, 8, 8, 1, 1, 1, 1, -9], k = 1
Output: [1]

Offset 1 was emitted four times, more than 8 at three, 3 at two and -9 at one.

Example 3

Input: readings = [10000, -10000, 5, -5], k = 4
Output: [10000, -10000, 5, -5]

Nothing repeats and k covers the whole distinct set, so all four offsets are reported.

Constraints

1 <= readings.length <= 10^5
-10^4 <= readings[i] <= 10^4
1 <= k <= the count of distinct values in readings
The set of k answers is guaranteed to be unambiguous

The values you return may be in any order.

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 frequent_offsets(readings: list[int], k: int) -> list[int]:

Java

public int[] frequentOffsets(int[] readings, int k)

The intended approach

Time O(n)Space O(n)

Tally each offset in a hash map on one pass. A tally can never exceed the length of the log, so file every offset into a bucket indexed by its tally, then read buckets from the highest index downward and stop after k values. No comparison sort is needed, so the whole thing stays linear.

Pythonentry point: frequent_offsets

def frequent_offsets(readings, k):
    tally = {}
    for value in readings:
        tally[value] = tally.get(value, 0) + 1
    buckets = [[] for _ in range(len(readings) + 1)]
    for value, hits in tally.items():
        buckets[hits].append(value)
    picked = []
    for hits in range(len(readings), 0, -1):
        for value in buckets[hits]:
            picked.append(value)
            if len(picked) == k:
                return picked
    return picked

The obvious version

Count each distinct offset by rescanning the whole log for it, then run k selection passes that each sweep the tallies for the largest one still unpicked. O(n^2).

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_frequent_offsets(readings, k):
    distinct = []
    for value in readings:
        if value not in distinct:
            distinct.append(value)
    tallies = []
    for value in distinct:
        hits = 0
        for other in readings:
            if other == value:
                hits += 1
        tallies.append([value, hits])
    picked = []
    for _ in range(k):
        best = None
        for value, hits in tallies:
            if value in picked:
                continue
            if best is None or hits > best[1]:
                best = [value, hits]
        picked.append(best[0])
    return picked

How this problem was checked

Reference solution executed against 8 test cases — 8 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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