0450EasyArrayHash Table

Distinct Tallies In A Core Log

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Trains the technique from

LeetCode 1207Unique Number of Occurrences

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 drilling crew tags every core sample it lifts with the signed offset of the level it came from, in decimetres above the survey datum. Levels below the datum carry a negative offset, and offsets lists one tag per sample in the order the samples were lifted.

The crew calls a log well spread when no two different offsets were sampled the same number of times. An offset that appears once and another that also appears once spoils that, while an offset sampled once alongside one sampled four times does not.

Return true if the log is well spread and false otherwise. Note that -40 and 40 are different offsets.

Examples

Example 1

Input: offsets = [-40, 7, -40, 7, 65]
Output: false

Offset -40 was sampled twice, offset 7 was sampled twice as well, so two different offsets share a tally.

Example 2

Input: offsets = [-5, -5, -5, 62, 91, 62, 14]
Output: false

The tallies are three samples at -5, two at 62 and one each at 91 and 14, and the two singles clash.

Example 3

Input: offsets = [-1000, -1000, -1000, -1000]
Output: true

Only one offset appears in the log, so there is no second tally for it to clash with.

Constraints

1 <= offsets.length <= 1000
-1000 <= offsets[i] <= 1000

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 all_tallies_differ(offsets: list[int]) -> bool:

Java

public boolean allTalliesDiffer(int[] offsets)

The intended approach

Time O(n) for n samples.Space O(d) for d distinct offsets, held in the tally map and the set of seen tallies.

Tally the samples per offset in one pass with a map keyed by the signed offset. Then read the tallies off and keep the ones already seen in a set, answering false the moment a tally turns up twice.

Pythonentry point: all_tallies_differ

def all_tallies_differ(offsets):
    # One pass to tally each offset, one pass to see whether any tally repeats.
    # O(n) time, O(d) extra space for d distinct offsets.
    tally = {}
    for off in offsets:
        tally[off] = tally.get(off, 0) + 1
    seen = set()
    for count in tally.values():
        if count in seen:
            return False
        seen.add(count)
    return True

The obvious version

Find the first appearance of every offset by scanning back over the log, count each one with a full sweep, then compare every pair of counts. 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_all_tallies_differ(offsets):
    firsts = []
    for i in range(len(offsets)):
        fresh = True
        for j in range(i):
            if offsets[j] == offsets[i]:
                fresh = False
                break
        if fresh:
            firsts.append(i)
    counts = []
    for i in firsts:
        c = 0
        for v in offsets:
            if v == offsets[i]:
                c += 1
        counts.append(c)
    for i in range(len(counts)):
        for j in range(i + 1, len(counts)):
            if counts[i] == counts[j]:
                return False
    return True

How this problem was checked

Reference solution executed against 12 test cases — 12 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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