1040MediumArrayHash TableSliding Window

Stretches With Enough Matching Pairs

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Trains the technique from

LeetCode 2537Count the Number of Good Subarrays

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 log holds the readings readings. A matching pair inside a stretch of neighbouring readings is any two positions of that stretch holding the same reading.

A stretch is rich when it holds at least k matching pairs.

Return how many stretches are rich.

Examples

Example 1

Input: readings = [3, 1, 4, 3, 2, 2, 4], k = 2
Output: 4

Four stretches hold at least two matching pairs. Each of them reaches from somewhere at or before the second reading out to the last, picking up the repeated 3, the repeated 4 and the repeated 2 along the way.

Example 2

Input: readings = [1, 1, 1, 1, 1], k = 10
Output: 1

Five readings all alike give ten matching pairs, exactly the number wanted, and only the whole log manages it.

Example 3

Input: readings = [1, 2, 3], k = 1
Output: 0

No two readings are alike, so no stretch holds a matching pair at all.

Constraints

1 <= readings.length <= 10^5
1 <= readings[i] <= 10^9
1 <= k <= 10^9

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

Java

public long countGood(int[] readings, int k)

The intended approach

Time O(n) over the log's length.Space O(d) for the tally over distinct readings.

Bringing a reading into a window raises its matching pairs by however many copies of that reading the window already held, since the newcomer pairs with each of them. Dropping a reading off the left lowers the pairs by however many copies are left once it has gone. So the pair count is kept up to date in a step, with a tally per distinct reading. A window's pair count only grows as its right end moves out and only shrinks as its left end moves in. So for each right end, push the left end in as far as it will go while the window still holds at least `k` pairs, and then every left end at or before that place also gives a rich stretch. That means the right end contributes exactly as many stretches as there are places the left end could have started. Both ends only ever move rightwards, so each reading enters and leaves the window once.

Pythonentry point: count_good

def count_good(readings, k):
    tally = {}
    pairs = 0
    total = 0
    lo = 0
    for hi in range(len(readings)):
        value = readings[hi]
        pairs += tally.get(value, 0)
        tally[value] = tally.get(value, 0) + 1
        while pairs >= k:
            gone = readings[lo]
            tally[gone] -= 1
            pairs -= tally[gone]
            lo += 1
        total += lo
    return total

The obvious version

Take every starting place and grow the stretch one reading at a time, keeping a tally as it goes and raising the pair count by the copies already present, then count the stretch whenever the pairs reach `k`. Costs the log's length squared rather than one window sliding along.

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_count_good(readings, k):
    n = len(readings)
    total = 0
    for start in range(n):
        tally = {}
        pairs = 0
        for end in range(start, n):
            value = readings[end]
            pairs += tally.get(value, 0)
            tally[value] = tally.get(value, 0) + 1
            if pairs >= k:
                total += 1
    return total

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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