LeetCode #3409 — MEDIUM

Longest Subsequence With Decreasing Adjacent Difference

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of integers nums.

Your task is to find the length of the longest subsequence seq of nums, such that the absolute differences between consecutive elements form a non-increasing sequence of integers. In other words, for a subsequence seq₀, seq₁, seq₂, ..., seq_m of nums, |seq₁ - seq₀| >= |seq₂ - seq₁| >= ... >= |seq_m - seq_{m - 1}|.

Return the length of such a subsequence.

Example 1:

Input: nums = [16,6,3]

Output: 3

Explanation:

The longest subsequence is [16, 6, 3] with the absolute adjacent differences [10, 3].

Example 2:

Input: nums = [6,5,3,4,2,1]

Output: 4

Explanation:

The longest subsequence is [6, 4, 2, 1] with the absolute adjacent differences [2, 2, 1].

Example 3:

Input: nums = [10,20,10,19,10,20]

Output: 5

Explanation:

The longest subsequence is [10, 20, 10, 19, 10] with the absolute adjacent differences [10, 10, 9, 9].

Constraints:

2 <= nums.length <= 10⁴
1 <= nums[i] <= 300

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums. Your task is to find the length of the longest subsequence seq of nums, such that the absolute differences between consecutive elements form a non-increasing sequence of integers. In other words, for a subsequence seq0, seq1, seq2, ..., seqm of nums, |seq1 - seq0| >= |seq2 - seq1| >= ... >= |seqm - seqm - 1|. Return the length of such a subsequence.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[16,6,3]

Example 2

[6,5,3,4,2,1]

Example 3

[10,20,10,19,10,20]

Core Insight

What unlocks the optimal approach

Use dynamic programming.
Store the maximum answer for each index and every possible difference.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #3409: Longest Subsequence With Decreasing Adjacent Difference
class Solution {
  public int longestSubsequence(int[] nums) {
    final int mx = Arrays.stream(nums).max().getAsInt();
    // dp[num][diff] := the length of the longest subsequence ending in `num`
    // s.t. the last absolute difference between consecutive elements is `diff`
    int[][] dp = new int[mx + 1][mx + 1];

    for (final int num : nums) {
      for (int prev = 1; prev <= mx; ++prev) {
        final int diff = Math.abs(num - prev);
        dp[num][diff] = Math.max(dp[num][diff], dp[prev][diff] + 1);
      }
      // dp[num][diff] := max(dp[num][j]), where j >= diff
      for (int j = mx - 1; j >= 0; --j)
        dp[num][j] = Math.max(dp[num][j], dp[num][j + 1]);
    }

    return Arrays.stream(dp).mapToInt(row -> row[0]).max().getAsInt();
  }
}

// Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
package main

import (
	"math"
	"slices"
)

// https://space.bilibili.com/206214
func longestSubsequence(nums []int) (ans int) {
	mx := slices.Max(nums)
	maxD := mx - slices.Min(nums)
	f := make([][]int, mx+1)
	for i := range f {
		f[i] = make([]int, maxD+1)
		for j := range f[i] {
			f[i][j] = math.MinInt
		}
	}

	for _, x := range nums {
		fx := 1
		for j := maxD; j >= 0; j-- {
			if x-j >= 0 {
				fx = max(fx, f[x-j][j]+1)
			}
			if x+j <= mx {
				fx = max(fx, f[x+j][j]+1)
			}
			f[x][j] = fx
			ans = max(ans, fx)
		}
	}
	return
}

func longestSubsequence2(nums []int) (ans int) {
	mx := slices.Max(nums)
	maxD := mx - slices.Min(nums)
	f := make([][]int, len(nums))
	for i := range f {
		f[i] = make([]int, maxD+2)
	}
	last := make([]int, mx+1)
	for i := range last {
		last[i] = -1
	}

	for i, x := range nums {
		for j := maxD; j >= 0; j-- {
			f[i][j] = max(f[i][j+1], 1)
			if x-j >= 0 && last[x-j] >= 0 {
				f[i][j] = max(f[i][j], f[last[x-j]][j]+1)
			}
			if x+j <= mx && last[x+j] >= 0 {
				f[i][j] = max(f[i][j], f[last[x+j]][j]+1)
			}
			ans = max(ans, f[i][j])
		}
		last[x] = i
	}
	return
}

# Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
class Solution:
  def longestSubsequence(self, nums: list[int]) -> int:
    mx = max(nums)
    # dp[num][diff] := the length of the longest subsequence ending in `num`
    # s.t. the last absolute difference between consecutive elements is `diff`
    dp = [[0] * (mx + 1) for _ in range(mx + 1)]

    for num in nums:
      for prev in range(1, mx + 1):
        diff = abs(num - prev)
        dp[num][diff] = max(dp[num][diff], dp[prev][diff] + 1)
      # dp[num][diff] := max(dp[num][j]) for j >= diff
      for j in range(mx - 1, -1, -1):
        dp[num][j] = max(dp[num][j], dp[num][j + 1])

    return max(map(max, dp))

// Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
// class Solution {
//   public int longestSubsequence(int[] nums) {
//     final int mx = Arrays.stream(nums).max().getAsInt();
//     // dp[num][diff] := the length of the longest subsequence ending in `num`
//     // s.t. the last absolute difference between consecutive elements is `diff`
//     int[][] dp = new int[mx + 1][mx + 1];
// 
//     for (final int num : nums) {
//       for (int prev = 1; prev <= mx; ++prev) {
//         final int diff = Math.abs(num - prev);
//         dp[num][diff] = Math.max(dp[num][diff], dp[prev][diff] + 1);
//       }
//       // dp[num][diff] := max(dp[num][j]), where j >= diff
//       for (int j = mx - 1; j >= 0; --j)
//         dp[num][j] = Math.max(dp[num][j], dp[num][j + 1]);
//     }
// 
//     return Arrays.stream(dp).mapToInt(row -> row[0]).max().getAsInt();
//   }
// }

// Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3409: Longest Subsequence With Decreasing Adjacent Difference
// class Solution {
//   public int longestSubsequence(int[] nums) {
//     final int mx = Arrays.stream(nums).max().getAsInt();
//     // dp[num][diff] := the length of the longest subsequence ending in `num`
//     // s.t. the last absolute difference between consecutive elements is `diff`
//     int[][] dp = new int[mx + 1][mx + 1];
// 
//     for (final int num : nums) {
//       for (int prev = 1; prev <= mx; ++prev) {
//         final int diff = Math.abs(num - prev);
//         dp[num][diff] = Math.max(dp[num][diff], dp[prev][diff] + 1);
//       }
//       // dp[num][diff] := max(dp[num][j]), where j >= diff
//       for (int j = mx - 1; j >= 0; --j)
//         dp[num][j] = Math.max(dp[num][j], dp[num][j + 1]);
//     }
// 
//     return Arrays.stream(dp).mapToInt(row -> row[0]).max().getAsInt();
//   }
// }

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.

Longest Subsequence With Decreasing Adjacent Difference

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

State misses one required dimension