LeetCode 1027: Longest Arithmetic Subsequence Solution

Master LeetCode problem 1027 (Longest Arithmetic Subsequence), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1027. Longest Arithmetic Subsequence

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will create a 2D dp array where dp[i][diff] represents the length of the longest arithmetic subsequence ending at index i with a common difference of diff. We iterate through each pair of elements (nums[i] and nums[j], where j < i) and update the dp array accordingly. Finally, we return the maximum value in the dp array as the result.

Algorithm:

  1. Initialize a 2D dp array of size nums.length x nums.length with all values set to 2 (since we have at least two elements in any subsequence).
  2. Initialize a variable maxLength to store the result.
  3. Iterate over each pair of elements (nums[i] and nums[j]) where j < i:
    • Calculate the common difference diff = nums[i] - nums[j].
    • Update dp[i][diff] = max(dp[i][diff], dp[j][diff] + 1).
    • Update maxLength = max(maxLength, dp[i][diff]).
  4. Return maxLength as the final result.

Time Complexity: O(n^2), where n is the length of the input array nums. Space Complexity: O(n^2), for the 2D dp array.

:

Solution Code

class Solution {
    public int longestArithSeqLength(int[] nums) {
        int n = nums.length;
        int[][] dp = new int[n][n];
        int maxLength = 2;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                int diff = nums[i] - nums[j];
                dp[i][diff] = Math.max(dp[i][diff], dp[j][diff] + 1);
                maxLength = Math.max(maxLength, dp[i][diff]);
            }
        }
        
        return maxLength;
    }
}

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