1111EasyArrayHash TableCounting

The Largest Self-Matching Token

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Trains the technique from

LeetCode 1394Find Lucky Integer in an Array

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 tray holds tokens whose face values are listed in tokens. A value is matched when the tray holds exactly that many tokens of it, so 3 is matched when exactly three tokens show 3.

Return the largest matched value, or -1 when the tray holds no matched value at all.

Examples

Example 1

Input: tokens = [2, 2, 2]
Output: -1

Three tokens show 2, which is one too many for 2 to be matched, and no other value appears at all.

Example 2

Input: tokens = [4, 4, 4, 4, 5, 5, 5, 5, 5]
Output: 5

Four tokens show 4 and five show 5, so both values are matched and the larger is reported.

Example 3

Input: tokens = [1]
Output: 1

A lone token showing 1 is matched, since 1 turns up exactly once.

Constraints

1 <= tokens.length <= 500
1 <= tokens[i] <= 500

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 find_lucky(tokens: list[int]) -> int:

Java

public int findLucky(int[] tokens)

The intended approach

Time O(n) over the tokens.Space O(n) for the tally.

Whether a value is matched depends on how often it turns up, so one pass building a tally of value against count settles every value at once. Then walk the tally and keep the largest value whose count equals it, starting from a sentinel below every possible value so that finding nothing reports back correctly. Walking the tray in its own order and stopping at the first match would be wrong, since the answer asks for the largest matched value and the tray is in no particular order. Nor is it enough for a value to turn up at least that often: a value appearing more times than itself is not matched.

Pythonentry point: find_lucky

def find_lucky(tokens):
    tally = {}
    for value in tokens:
        tally[value] = tally.get(value, 0) + 1
    best = -1
    for value in tally:
        if tally[value] == value and value > best:
            best = value
    return best

The obvious version

For each token in the tray, sweep the whole tray counting how many share its value, and keep the largest value whose count comes out equal to it. Costs the token count squared rather than one 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_find_lucky(tokens):
    best = -1
    for value in tokens:
        count = 0
        for other in tokens:
            if other == value:
                count += 1
        if count == value and value > best:
            best = value
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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