LeetCode #1027 — MEDIUM

Longest Arithmetic Subsequence

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

Solve on LeetCode

The Problem

Problem Statement

Given an array nums of integers, return the length of the longest arithmetic subsequence in nums.

Note that:

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.
A sequence seq is arithmetic if seq[i + 1] - seq[i] are all the same value (for 0 <= i < seq.length - 1).

Example 1:

Input: nums = [3,6,9,12]
Output: 4
Explanation:  The whole array is an arithmetic sequence with steps of length = 3.

Example 2:

Input: nums = [9,4,7,2,10]
Output: 3
Explanation:  The longest arithmetic subsequence is [4,7,10].

Example 3:

Input: nums = [20,1,15,3,10,5,8]
Output: 4
Explanation:  The longest arithmetic subsequence is [20,15,10,5].

Constraints:

2 <= nums.length <= 1000
0 <= nums[i] <= 500

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: Given an array nums of integers, return the length of the longest arithmetic subsequence in nums. Note that: 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. A sequence seq is arithmetic if seq[i + 1] - seq[i] are all the same value (for 0 <= i < seq.length - 1).

Baseline thinking

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

Pattern signal: Array · Hash Map · Binary Search · Dynamic Programming

Example 1

[3,6,9,12]

Example 2

[9,4,7,2,10]

Example 3

[20,1,15,3,10,5,8]

Related Problems

Destroy Sequential Targets (destroy-sequential-targets)

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #1027: Longest Arithmetic Subsequence
class Solution {
    public int longestArithSeqLength(int[] nums) {
        int n = nums.length;
        int ans = 0;
        int[][] f = new int[n][1001];
        for (int i = 1; i < n; ++i) {
            for (int k = 0; k < i; ++k) {
                int j = nums[i] - nums[k] + 500;
                f[i][j] = Math.max(f[i][j], f[k][j] + 1);
                ans = Math.max(ans, f[i][j]);
            }
        }
        return ans + 1;
    }
}

// Accepted solution for LeetCode #1027: Longest Arithmetic Subsequence
func longestArithSeqLength(nums []int) int {
	n := len(nums)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, 1001)
	}
	ans := 0
	for i := 1; i < n; i++ {
		for k := 0; k < i; k++ {
			j := nums[i] - nums[k] + 500
			f[i][j] = max(f[i][j], f[k][j]+1)
			ans = max(ans, f[i][j])
		}
	}
	return ans + 1
}

# Accepted solution for LeetCode #1027: Longest Arithmetic Subsequence
class Solution:
    def longestArithSeqLength(self, nums: List[int]) -> int:
        n = len(nums)
        f = [[1] * 1001 for _ in range(n)]
        ans = 0
        for i in range(1, n):
            for k in range(i):
                j = nums[i] - nums[k] + 500
                f[i][j] = max(f[i][j], f[k][j] + 1)
                ans = max(ans, f[i][j])
        return ans

// Accepted solution for LeetCode #1027: Longest Arithmetic Subsequence
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn longest_arith_seq_length(a: Vec<i32>) -> i32 {
        let mut res = 0;
        let n = a.len();
        let mut dp: Vec<HashMap<i32, i32>> = vec![HashMap::new(); n];
        for i in 0..n {
            for j in 0..i {
                let diff = a[i] - a[j];
                let len_j = *dp[j].entry(diff).or_default();
                let len_i = dp[i].entry(diff).or_default();
                *len_i = len_j + 1;
                res = res.max(*len_i);
            }
        }
        res + 1
    }
}

#[test]
fn test() {
    let a = vec![3, 6, 9, 12];
    let res = 4;
    assert_eq!(Solution::longest_arith_seq_length(a), res);
    let a = vec![9, 4, 7, 2, 10];
    let res = 3;
    assert_eq!(Solution::longest_arith_seq_length(a), res);
    let a = vec![20, 1, 15, 3, 10, 5, 8];
    let res = 4;
    assert_eq!(Solution::longest_arith_seq_length(a), res);
}

// Accepted solution for LeetCode #1027: Longest Arithmetic Subsequence
function longestArithSeqLength(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    const f: number[][] = Array.from({ length: n }, () => new Array(1001).fill(0));
    for (let i = 1; i < n; ++i) {
        for (let k = 0; k < i; ++k) {
            const j = nums[i] - nums[k] + 500;
            f[i][j] = Math.max(f[i][j], f[k][j] + 1);
            ans = Math.max(ans, f[i][j]);
        }
    }
    return ans + 1;
}

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 × (d + n)

Space

O(n × d)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

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.