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2407. Longest Increasing Subsequence II

Description

You are given an integer array nums and an integer k.

Find the longest subsequence of nums that meets the following requirements:

  • The subsequence is strictly increasing and
  • The difference between adjacent elements in the subsequence is at most k.

Return the length of the longest subsequence that meets the requirements.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

 

Example 1:

Input: nums = [4,2,1,4,3,4,5,8,15], k = 3
Output: 5
Explanation:
The longest subsequence that meets the requirements is [1,3,4,5,8].
The subsequence has a length of 5, so we return 5.
Note that the subsequence [1,3,4,5,8,15] does not meet the requirements because 15 - 8 = 7 is larger than 3.

Example 2:

Input: nums = [7,4,5,1,8,12,4,7], k = 5
Output: 4
Explanation:
The longest subsequence that meets the requirements is [4,5,8,12].
The subsequence has a length of 4, so we return 4.

Example 3:

Input: nums = [1,5], k = 1
Output: 1
Explanation:
The longest subsequence that meets the requirements is [1].
The subsequence has a length of 1, so we return 1.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i], k <= 105

Solutions

Solution 1: Segment Tree

We assume that $f[v]$ represents the length of the longest increasing subsequence ending with the number $v$.

We traverse each element $v$ in the array $nums$, with the state transition equation: $f[v] = \max(f[v], f[x])$, where the range of $x$ is $[v-k, v-1]$.

Therefore, we need a data structure to maintain the maximum value of the interval. It is not difficult to think of using a segment tree.

The segment tree divides the entire interval into multiple discontinuous subintervals, and the number of subintervals does not exceed $log(width)$. To update the value of an element, only $log(width)$ intervals need to be updated, and these intervals are all contained in a large interval that contains the element.

  • Each node of the segment tree represents an interval;
  • The segment tree has a unique root node, which represents the entire statistical range, such as $[1,N]$;
  • Each leaf node of the segment tree represents an elementary interval of length $1$, $[x, x]$;
  • For each internal node $[l,r]$, its left child is $[l,mid]$, and the right child is $[mid+1,r]$, where $mid = \left \lfloor \frac{l+r}{2} \right \rfloor$.

For this problem, the information maintained by the segment tree node is the maximum value within the interval range.

The time complexity is $O(n \times \log n)$, where $n$ is the length of the array $nums$.

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class Node:
    def __init__(self):
        self.l = 0
        self.r = 0
        self.v = 0


class SegmentTree:
    def __init__(self, n):
        self.tr = [Node() for _ in range(4 * n)]
        self.build(1, 1, n)

    def build(self, u, l, r):
        self.tr[u].l = l
        self.tr[u].r = r
        if l == r:
            return
        mid = (l + r) >> 1
        self.build(u << 1, l, mid)
        self.build(u << 1 | 1, mid + 1, r)

    def modify(self, u, x, v):
        if self.tr[u].l == x and self.tr[u].r == x:
            self.tr[u].v = v
            return
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        if x <= mid:
            self.modify(u << 1, x, v)
        else:
            self.modify(u << 1 | 1, x, v)
        self.pushup(u)

    def pushup(self, u):
        self.tr[u].v = max(self.tr[u << 1].v, self.tr[u << 1 | 1].v)

    def query(self, u, l, r):
        if self.tr[u].l >= l and self.tr[u].r <= r:
            return self.tr[u].v
        mid = (self.tr[u].l + self.tr[u].r) >> 1
        v = 0
        if l <= mid:
            v = self.query(u << 1, l, r)
        if r > mid:
            v = max(v, self.query(u << 1 | 1, l, r))
        return v


class Solution:
    def lengthOfLIS(self, nums: List[int], k: int) -> int:
        tree = SegmentTree(max(nums))
        ans = 1
        for v in nums:
            t = tree.query(1, v - k, v - 1) + 1
            ans = max(ans, t)
            tree.modify(1, v, t)
        return ans

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class Solution {
    public int lengthOfLIS(int[] nums, int k) {
        int mx = nums[0];
        for (int