0306MediumArrayGreedyLongest Increasing Subsequence

Three Rising Gauge Marks

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Trains the technique from

LeetCode 334Increasing Triplet Subsequence

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 river gauge writes one reading per hour into a log. readings[t] is the water level at hour t, measured against the datum mark, so it may be negative when the level sits below the mark.

The flood office wants to know whether the log contains a rising trio: three hours i < j < k with readings[i] < readings[j] < readings[k]. The three hours do not have to be next to each other, but they must appear in that order in the log, and each step must be a strict increase.

Return true if the log contains a rising trio and false otherwise. Your solution should run in O(n) time and use O(1) extra space.

Examples

Example 1

Input: readings = [9, 4, -2, 5, 1, 7]
Output: true

Hours 2, 3 and 5 hold -2, 5 and 7. Those hours are in increasing order and -2 < 5 < 7, so the log has a rising trio.

Example 2

Input: readings = [8, 8, 3, 3, -1, -1]
Output: false

No reading is ever followed later by a larger one, so not even a rising pair exists, let alone a trio.

Example 3

Input: readings = [6, 1, 7, 2, 8]
Output: true

Hours 1, 3 and 4 hold 1, 2 and 8, and 1 < 2 < 8.

Example 4

Input: readings = [4, 5, 1, 2]
Output: false

The rising pairs available are (4, 5) and (1, 2), and neither has a third larger reading after it, so there is no trio.

Example 5

Input: readings = [5, 5, 5, 5]
Output: false

Every reading equals the next, and equal readings are not a strict increase.

Constraints

1 <= readings.length <= 5 * 10^5
-2^31 <= readings[i] <= 2^31 - 1

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 has_rising_trio(readings: list[int]) -> bool:

Java

public boolean hasRisingTrio(int[] readings)

The intended approach

Time O(n)Space O(1)

Keep two running values while scanning left to right: `low`, the smallest reading seen so far, and `mid`, the smallest reading that already has a strictly smaller reading somewhere before it. A reading at or below `low` replaces `low`; a reading at or below `mid` replaces `mid`; anything larger than both completes a trio. Lowering `low` after `mid` is set is safe, because `mid` still records that some earlier reading was below it.

Pythonentry point: has_rising_trio

def has_rising_trio(readings):
    # O(n) time, single pass. O(1) space: two scalars, nothing rebuilt.
    low = None
    mid = None
    for v in readings:
        if low is None or v <= low:
            low = v
        elif mid is None or v <= mid:
            mid = v
        else:
            return True
    return False

The obvious version

Test every ordered triple of hours with three nested loops and report whether any of them rises twice.

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_has_rising_trio(readings):
    n = len(readings)
    for i in range(n):
        for j in range(i + 1, n):
            if readings[j] <= readings[i]:
                continue
            for k in range(j + 1, n):
                if readings[k] > readings[j]:
                    return True
    return False

How this problem was checked

Reference solution executed against 14 test cases — 14 passed
5 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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