0464MediumArrayDynamic ProgrammingBinary Indexed TreeSegment TreeLongest Increasing Subsequence

Counting The Longest Rising Gauge Runs

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Trains the technique from

LeetCode 673Number of Longest 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 reports its level against a datum, so a reading may be negative. marks holds one reading per hour, earliest hour first.

A rising run is a pick of one or more readings, kept in the order the hours were logged, in which each reading picked is strictly above the one picked before it. The hours picked need not be consecutive. Two runs are different when they pick different hours, even if the readings they pick happen to match.

Return how many rising runs are as long as the longest rising run in the log.

Examples

Example 1

Input: marks = [-14, 62, 5, 90, 41]
Output: 3

The longest rising runs pick three hours: -14, 62, 90 and -14, 5, 90 and -14, 5, 41.

Example 2

Input: marks = [77, 77, 77]
Output: 3

No reading is strictly above another, so the longest run picks a single hour, and each of the three hours gives one such run.

Example 3

Input: marks = [-8, -20, 46, -3, 46]
Output: 2

The longest runs pick three hours: -20 then -3 then the later 46, and -8 then -3 then that same 46.

Constraints

1 <= marks.length <= 2000
-10^6 <= marks[i] <= 10^6
The answer is guaranteed to fit in a signed 32-bit integer.

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_rise_count(marks: list[int]) -> int:

Java

public int longestRiseCount(int[] marks)

The intended approach

Time O(n^2) for n hours, since every ordered pair of hours is weighed once.Space O(n) for the two figures per hour.

Give every hour two figures: the length of the longest rising run that finishes at that hour, and how many runs of that length finish there. Both come from the earlier hours whose reading is strictly lower - one that beats the current length replaces both figures, one that only matches it adds its own count on. The answer sums the counts of every hour whose length equals the largest length found.

Pythonentry point: longest_rise_count

def longest_rise_count(marks):
    # For each reading keep two figures: the longest rising pick that ends there
    # and how many such picks there are. A strictly earlier, strictly lower
    # reading either extends the record or matches it, and the two cases are
    # handled differently.
    # O(n^2) time, O(n) space.
    n = len(marks)
    span = [1] * n
    ways = [1] * n
    for j in range(n):
        mj = marks[j]
        for i in range(j):
            if marks[i] < mj:
                reach = span[i] + 1
                if reach > span[j]:
                    span[j] = reach
                    ways[j] = ways[i]
                elif reach == span[j]:
                    ways[j] += ways[i]
    top = max(span)
    total = 0
    for j in range(n):
        if span[j] == top:
            total += ways[j]
    return total

The obvious version

Enumerate every pick of hours in log order, keep only the strictly rising ones, and tally how many share the greatest length seen. O(2^n * n).

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_rise_count(marks):
    n = len(marks)
    best = 0
    tally = 0
    for mask in range(1, 1 << n):
        pick = [marks[i] for i in range(n) if mask >> i & 1]
        ok = True
        for t in range(len(pick) - 1):
            if pick[t] >= pick[t + 1]:
                ok = False
                break
        if not ok:
            continue
        if len(pick) > best:
            best = len(pick)
            tally = 1
        elif len(pick) == best:
            tally += 1
    return tally

How this problem was checked

Reference solution executed against 14 test cases — 14 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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