LeetCode 2370: Longest Ideal Subsequence Solution

Master LeetCode problem 2370 (Longest Ideal 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.

2370. Longest Ideal Subsequence

Medium

Hash Table String Dynamic Programming

Problem Explanation

Explanation

To find the length of the longest ideal subsequence, we can use a dynamic programming approach. We can define a 2D array dp where dp[i][j] represents the length of the longest ideal subsequence ending at index i with a difference of j between the last two characters. We can then iterate over the string s and update the dp array based on the current character and the previous characters.

Algorithm:

Initialize a 2D array dp of size n x 26 where n is the length of the string s and 26 represents the number of lowercase letters.
Initialize all values in dp with 0.
Iterate over each character in the string s:
- For each character c at index i, consider all possible characters prev that can precede c such that the difference between c and prev is less than or equal to k.
- Update the dp[i][c - 'a'] with the maximum value of dp[prevIndex][prevChar - 'a'] + 1 for all valid prev characters.
Find the maximum value in the last row of dp to get the length of the longest ideal subsequence.

Time Complexity: The time complexity of this algorithm is O(n * k) where n is the length of the string s.

Space Complexity: The space complexity of this algorithm is O(n * 26) = O(n) where n is the length of the string s.

Solutions

Solution Code

class Solution {
    public int longestIdealSubsequence(String s, int k) {
        int n = s.length();
        int[][] dp = new int[n][26];

        for (int i = 0; i < n; i++) {
            char c = s.charAt(i);
            for (char prev = 'a'; prev <= 'z'; prev++) {
                int prevIndex = i - 1;
                while (prevIndex >= 0 && Math.abs(c - prev) <= k) {
                    dp[i][c - 'a'] = Math.max(dp[i][c - 'a'], dp[prevIndex][prev - 'a'] + 1);
                    prevIndex--;
                }
            }
        }

        int maxLen = 0;
        for (int j = 0; j < 26; j++) {
            maxLen = Math.max(maxLen, dp[n - 1][j]);
        }

        return maxLen;
    }
}

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