README.md

1027. Longest Arithmetic Subsequence

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn longest_arith_seq_length(nums: Vec<i32>) -> i32 {
        let max_num = *nums.iter().max().unwrap();
        let mut dp = vec![vec![0; max_num as usize * 2 + 1]; max_num as usize + 1];
        let mut ret = 2;

        for i in 0..nums.len() {
            for diff in -max_num..=max_num {
                if nums[i] - diff < 0 || nums[i] - diff > max_num {
                    dp[nums[i] as usize][(diff + max_num) as usize] = 1;
                } else {
                    dp[nums[i] as usize][(diff + max_num) as usize] =
                        dp[(nums[i] - diff) as usize][(diff + max_num) as usize] + 1;
                }
                ret = ret.max(dp[nums[i] as usize][(diff + max_num) as usize]);
            }
        }

        ret
    }
}