0901HardArrayDivide and ConquerDynamic ProgrammingBinary Indexed TreeSegment TreeQueueMonotonic Queue

Longest Rising Run Under a Step Cap

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Trains the technique from

LeetCode 2407Longest Increasing Subsequence 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.

Gauge readings are given as nums. A climb picks readings from the list, keeping their order, so that every reading is strictly larger than the one before it while the rise from one to the next is never more than k.

Return how many readings the longest climb holds.

Examples

Example 1

Input: nums = [13, 6, 19, 7, 21, 8, 25], k = 7
Output: 4

The climb 6, 7, 8 rises by one each time, and no longer climb keeps every step inside seven: reaching 13 then 19 is a step of six but 19 to 21 to 25 cannot be extended back far enough to beat three.

Example 2

Input: nums = [1, 3, 5, 7, 9], k = 2
Output: 5

Each reading is two above the last, which the cap just allows, so the whole list is one climb.

Example 3

Input: nums = [1, 3, 5, 7, 9], k = 1
Output: 1

Every step is two, which the cap forbids, so no climb holds more than one reading.

Constraints

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

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 length_of_l_i_s(nums: list[int], k: int) -> int:

Java

public int lengthOfLIS(int[] nums, int k)

The intended approach

Time O(n log V) for n readings over a value range V.Space O(V) for the tree over the value range.

Read the list once, working out for each reading the longest climb that ends on it. That length is one more than the best ending on any earlier reading whose value lies in the window from the current value minus `k` up to one below it, which is exactly the range the step cap allows. So the problem is a running series of largest-in-a-value-range questions with single-value updates in between. Index by value rather than by position and that pairing is what a segment tree over the value range gives: each reading costs one range query and one point update. Because the tree is keyed on value, only readings already passed have been written into it, so order is respected for free.

Pythonentry point: length_of_l_i_s

def length_of_l_i_s(nums, k):
    top = 0
    for v in nums:
        if v > top:
            top = v
    size = 1
    while size < top + 1:
        size *= 2
    tree = [0] * (2 * size)

    def largest(lo, hi):
        if lo > hi:
            return 0
        best = 0
        lo += size
        hi += size + 1
        while lo < hi:
            if lo & 1:
                if tree[lo] > best:
                    best = tree[lo]
                lo += 1
            if hi & 1:
                hi -= 1
                if tree[hi] > best:
                    best = tree[hi]
            lo >>= 1
            hi >>= 1
        return best

    def write(at, val):
        i = at + size
        if tree[i] >= val:
            return
        tree[i] = val
        i >>= 1
        while i:
            left = tree[2 * i]
            right = tree[2 * i + 1]
            m = left if left > right else right
            if tree[i] == m:
                break
            tree[i] = m
            i >>= 1

    best = 0
    for v in nums:
        lo = v - k
        if lo < 0:
            lo = 0
        run = largest(lo, v - 1) + 1
        if run > best:
            best = run
        write(v, run)
    return best

The obvious version

Work out the longest climb ending on each reading by looking back over every earlier reading and taking the best one the step cap allows. Quadratic in the reading count and it keeps no structure over the values at all.

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_length_of_l_i_s(nums, k):
    n = len(nums)
    ending = [1] * n
    for i in range(n):
        for j in range(i):
            if nums[j] < nums[i] and nums[i] - nums[j] <= k:
                if ending[j] + 1 > ending[i]:
                    ending[i] = ending[j] + 1
    best = 0
    for v in ending:
        if v > best:
            best = v
    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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