0926MediumArrayHash Table

Tightest Triple of Matching Readings

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Trains the technique from

LeetCode 3741Minimum Distance Between Three Equal Elements II

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.

Readings are given as nums.

Find three positions holding the same reading and return the span from the first of them to the last, which is the difference of those two positions. Return -1 when no reading appears three times.

Examples

Example 1

Input: nums = [4, 2, 4, 7, 4, 2, 2, 7, 7, 2]
Output: 4

The reading 2 sits at positions 1, 5, 6 and 9; the triple at 5, 6 and 9 spans 4, and the one at 1, 5 and 6 spans 5. The reading 4 sits at 0, 2 and 4, spanning 4, and the reading 7 at 3, 7 and 8, spanning 5. The tightest is 4.

Example 2

Input: nums = [1, 1, 1]
Output: 2

The only triple runs from position 0 to position 2.

Example 3

Input: nums = [1, 2, 3, 4, 5]
Output: -1

No reading appears even twice, let alone three times.

Constraints

1 <= nums.length <= 10^5
1 <= nums[i] <= nums.length

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 minimum_distance(nums: list[int]) -> int:

Java

public int minimumDistance(int[] nums)

The intended approach

Time O(n) for n readings, one pass.Space O(n) for the two remembered positions per reading.

For a fixed reading, the tightest triple among its positions must be three that sit next to each other in its own list of positions: skipping one only pushes the outer two further apart. So the answer looks at each reading's positions in order and takes the smallest gap between a position and the one two places later. That needs no storage per reading beyond the last two positions seen, so one pass suffices: for each reading, remember where it was seen the two times before, and whenever it turns up a third time or later, compare the span from two occurrences back against the best so far.

Pythonentry point: minimum_distance

def minimum_distance(nums):
    n = len(nums)
    first_back = {}
    second_back = {}
    best = -1
    for i in range(n):
        v = nums[i]
        if v in second_back:
            span = i - second_back[v]
            if best < 0 or span < best:
                best = span
        second_back[v] = first_back.get(v)
        first_back[v] = i
        if second_back[v] is None:
            del second_back[v]
    return best

The obvious version

Try every three positions, check that they hold the same reading, and keep the smallest span. Cubic in the reading count and it groups nothing.

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_minimum_distance(nums):
    n = len(nums)
    best = -1
    for i in range(n):
        for j in range(i + 1, n):
            if nums[j] != nums[i]:
                continue
            for k in range(j + 1, n):
                if nums[k] != nums[i]:
                    continue
                span = k - i
                if best < 0 or span < best:
                    best = span
    return best

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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