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1027. Longest Arithmetic Subsequence 👍

  • Time: $O(n^2)$
  • Space: $O(n^2)$
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class Solution {
 public:
  int longestArithSeqLength(vector<int>& nums) {
    const int n = nums.size();
    int ans = 0;
    // dp[i][k] := the length of the longest arithmetic subsequence of
    // nums[0..i] with k = diff + 500
    vector<vector<int>> dp(n, vector<int>(1001));

    for (int i = 0; i < n; ++i)
      for (int j = 0; j < i; ++j) {
        const int k = nums[i] - nums[j] + 500;
        dp[i][k] = max(2, dp[j][k] + 1);
        ans = max(ans, dp[i][k]);
      }

    return ans;
  }
};

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class Solution {
  public int longestArithSeqLength(int[] nums) {
    final int n = nums.length;
    int ans = 0;
    // dp[i][k] := the length of the longest arithmetic subsequence of nums[0..i]
    // with k = diff + 500
    int[][] dp = new int[n][1001];

    for (int i = 0; i < n; ++i)
      for (int j = 0; j < i; ++j) {
        final int k = nums[i] - nums[j] + 500;
        dp[i][k] = Math.max(2, dp[j][k] + 1);
        ans = Math.max(ans, dp[i][k]);
      }

    return ans;
  }
}

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class Solution:
  def longestArithSeqLength(self, nums: list[int]) -> int:
    n = len(nums)
    ans = 0
    # dp[i][k] := the length of the longest arithmetic subsequence of nums[0..i]
    # with k = diff + 500
    dp = [[0] * 1001 for _ in range(n)]

    for i in range(n):
      for j in range(i):
        k = nums[i] - nums[j] + 500
        dp[i][k] = max(2, dp[j][k] + 1)
        ans = max(ans, dp[i][k])

    return ans