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:
- 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).
- Initialize a variable
maxLengthto store the result. - 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]).
- 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;
}
}Try It Yourself
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