Leetcode #2407: Longest Increasing Subsequence II
In this guide, we solve Leetcode #2407 Longest Increasing Subsequence II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Binary Indexed Tree, Segment Tree, Queue, Array, Divide and Conquer, Dynamic Programming, Monotonic Queue
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
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
Complexity
The time complexity is , where is the length of the array . The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.