0491EasyArrayHash TableCounting

Peak Locker Tally

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Trains the technique from

LeetCode 3005Count Elements With Maximum 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 lost-property desk writes one ledger line for every item handed in, and each line records the number of the locker the item was stowed in. lockers[i] is the locker number written on line i, and the same locker number can appear on any number of lines.

Work out the largest number of lines any single locker number holds. Then add up the lines belonging to every locker number that holds exactly that many lines, and return that total.

Examples

Example 1

Input: lockers = [45, 78, 45, 91, 78, 12]
Output: 4

Lockers 45 and 78 hold two lines each, while 91 and 12 hold one each, so two lockers tie at the top and they account for four lines between them.

Example 2

Input: lockers = [63, 27, 84, 19]
Output: 4

Every locker number holds a single line, so all four numbers tie at the top and all four lines are counted.

Example 3

Input: lockers = [56, 56, 56, 31, 74, 31]
Output: 3

Locker 56 holds three lines and no other number holds three, so only its three lines are counted.

Constraints

1 <= lockers.length <= 100
1 <= lockers[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 peak_locker_lines(lockers: list[int]) -> int:

Java

public int peakLockerLines(int[] lockers)

The intended approach

Time O(n)Space O(1)

Because locker numbers never leave a small fixed range, a tally strip indexed by locker number holds every count. Fill it in one pass over the ledger, take the largest count in the strip, and then add together the counts that equal it, which is exactly the number of ledger lines those lockers account for.

Pythonentry point: peak_locker_lines

def peak_locker_lines(lockers):
    # Locker numbers are bounded, so a fixed tally strip is enough. One pass to
    # tally, one pass over the strip for the busiest count, one more to add up the
    # lines held by the lockers that reach it.
    # time O(n), space O(1) for the fixed 100-slot strip
    tally = [0] * 101
    for number in lockers:
        tally[number] += 1
    busiest = 0
    for count in tally:
        if count > busiest:
            busiest = count
    lines = 0
    for count in tally:
        if count == busiest:
            lines += count
    return lines

The obvious version

For each ledger line, rescan the whole ledger to see how many lines share its locker number. The largest of those per-line figures is the busiest count, and the answer is how many lines carry that figure. Quadratic and keeps no map at all.

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_peak_locker_lines(lockers):
    per_line = []
    for number in lockers:
        seen = 0
        for other in lockers:
            if other == number:
                seen += 1
        per_line.append(seen)
    busiest = 0
    for seen in per_line:
        if seen > busiest:
            busiest = seen
    total = 0
    for seen in per_line:
        if seen == busiest:
            total += 1
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 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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