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1027. Longest Arithmetic Subsequence

Description

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

Solution 1: Dynamic Programming

We define $f[i][j]$ as the maximum length of the arithmetic sequence ending with $nums[i]$ and having a common difference of $j$. Initially, $f[i][j]=1$, that is, each element itself is an arithmetic sequence of length $1$.

Since the common difference may be negative, and the maximum difference is $500$, we can uniformly add $500$ to the common difference, so the range of the common difference becomes $[0, 1000]$.

Considering $f[i]$, we can enumerate the previous element $nums[k]$ of $nums[i]$, then the common difference $j=nums[i]-nums[k]+500$, at this time $f[i][j]=\max(f[i][j], f[k][j]+1)$, then we update the answer $ans=\max(ans, f[i][j])$.

Finally, return the answer.

If initially $f[i][j]=0$, then we need to add $1$ to the answer when returning the answer.

The time complexity is $O(n \times (d + n))$, and the space complexity is $O(n \times d)$. Where $n$ and $d$ are the length of the array $nums$ and the difference between the maximum and minimum values in the array $nums$, respectively.

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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

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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;
    }
}

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class Solution {
public:
    int longestArithSeqLength(vector<int>& nums) {
        int n = nums.size();
        int f[n][1001];
        memset(f, 0, sizeof(f));
        int ans = 0;
        for (int i = 1; i < n; ++i) {
            for (int k = 0; k < i; ++k) {
                int 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;
    }
};

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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
}

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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;
}

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