0673EasyArray

Longest Rising Run of Gauge Readings

Tracked in this browser only

Trains the technique from

LeetCode 674Longest Continuous Increasing 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 water level per hour into readings, in the order the levels were taken. Levels are whole numbers and may be negative when the water is below the gauge's mark.

A rising run is a block of neighbouring entries readings[i] through readings[j] in which every entry is strictly greater than the entry just before it inside the block. A block of one entry counts as a rising run, and two equal entries side by side cannot be in the same run.

Return the number of entries in the longest rising run of readings.

Examples

Example 1

Input: readings = [4, 9, 11, 3, 6, 8, 20]
Output: 4

The block `3, 6, 8, 20` at the end holds four entries and each is strictly above the one before it, so it is a rising run of four.

Example 2

Input: readings = [15, 12, 9, 6]
Output: 1

Every entry is below the one before it, so no block of two entries rises and each single entry is a run of one.

Example 3

Input: readings = [9, 4, 6, 6, 7, 8, 9]
Output: 4

The two equal entries of `6` cannot share a run, so the run ending at the last entry starts at the second `6` and holds `6, 7, 8, 9`.

Constraints

1 <= readings.length <= 10^4
-10^9 <= readings[i] <= 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 longest_rising_run(readings: list[int]) -> int:

Java

public int longestRisingRun(int[] readings)

The intended approach

Time O(n): one pass over the readings.Space O(1).

Keep the length of the run that ends at the entry being read. Step through the readings once: if the current entry is strictly above its neighbour on the left the run grows by one, otherwise the current entry begins a run of length one. Track the largest run length seen, starting from one because a lone entry is already a run.

Pythonentry point: longest_rising_run

def longest_rising_run(readings):
    # one pass: the run either grows by one or restarts at the current entry
    best = 1
    run = 1
    for i in range(1, len(readings)):
        if readings[i] > readings[i - 1]:
            run += 1
            if run > best:
                best = run
        else:
            run = 1
    return best

The obvious version

Take every entry as a starting point and walk forward while each next entry is strictly larger, measuring how far the walk gets. Report the longest walk found over all starting points.

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_longest_rising_run(readings):
    n = len(readings)
    best = 0
    for start in range(n):
        end = start
        while end + 1 < n and readings[end + 1] > readings[end]:
            end += 1
        span = end - start + 1
        if span > best:
            best = span
    return best

How this problem was checked

Reference solution executed against 10 test cases — 10 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in