LeetCode #2407 — HARD

Longest Increasing Subsequence II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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.

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
Patterns Used
📊Dynamic Programming→📐Segment Tree→📈Monotonic Queue→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Segment Tree · Monotonic Queue

Example 1

[4,2,1,4,3,4,5,8,15]
3

Example 2

[7,4,5,1,8,12,4,7]
5

Example 3

[1,5]
1

Related Problems

  • Longest Increasing Subsequence (longest-increasing-subsequence)
  • Number of Longest Increasing Subsequence (number-of-longest-increasing-subsequence)
  • Longest Continuous Increasing Subsequence (longest-continuous-increasing-subsequence)
  • Longest Substring of One Repeating Character (longest-substring-of-one-repeating-character)
  • Booking Concert Tickets in Groups (booking-concert-tickets-in-groups)
Step 02

Core Insight

What unlocks the optimal approach

  • We can use dynamic programming. Let dp[i][val] be the answer using only the first i + 1 elements, and the last element in the subsequence is equal to val.
  • The only value that might change between dp[i - 1] and dp[i] are dp[i - 1][val] and dp[i][val].
  • Try using dp[i - 1] and the fact that the second last element in the subsequence has to fall within a range to calculate dp[i][val].
  • We can use a segment tree to find the maximum value in dp[i - 1] within a certain range.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2407: Longest Increasing Subsequence II
class Solution {
    public int lengthOfLIS(int[] nums, int k) {
        int mx = nums[0];
        for (int v : nums) {
            mx = Math.max(mx, v);
        }
        SegmentTree tree = new SegmentTree(mx);
        int ans = 0;
        for (int v : nums) {
            int t = tree.query(1, v - k, v - 1) + 1;
            ans = Math.max(ans, t);
            tree.modify(1, v, t);
        }
        return ans;
    }
}

class Node {
    int l;
    int r;
    int v;
}

class SegmentTree {
    private Node[] tr;

    public SegmentTree(int n) {
        tr = new Node[4 * n];
        for (int i = 0; i < tr.length; ++i) {
            tr[i] = new Node();
        }
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        tr[u].l = l;
        tr[u].r = r;
        if (l == r) {
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
    }

    public void modify(int u, int x, int v) {
        if (tr[u].l == x && tr[u].r == x) {
            tr[u].v = v;
            return;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        if (x <= mid) {
            modify(u << 1, x, v);
        } else {
            modify(u << 1 | 1, x, v);
        }
        pushup(u);
    }

    public void pushup(int u) {
        tr[u].v = Math.max(tr[u << 1].v, tr[u << 1 | 1].v);
    }

    public int query(int u, int l, int r) {
        if (tr[u].l >= l && tr[u].r <= r) {
            return tr[u].v;
        }
        int mid = (tr[u].l + tr[u].r) >> 1;
        int v = 0;
        if (l <= mid) {
            v = query(u << 1, l, r);
        }
        if (r > mid) {
            v = Math.max(v, query(u << 1 | 1, l, r));
        }
        return v;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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