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LeetCode 2030: Smallest K-Length Subsequence With Occurrences of a Letter Solution

Master LeetCode problem 2030 (Smallest K-Length Subsequence With Occurrences of a Letter), a hard 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.

2030. Smallest K-Length Subsequence With Occurrences of a Letter

Hard
StringStackGreedyMonotonic Stack

Problem Explanation

Explanation

To solve this problem, we can use a sliding window approach. We will iterate through the string s and maintain a window of size k. At each step, we will check if the current window contains the required number of occurrences of the given letter. If it does, we will update our result with the lexicographically smallest subsequence found so far.

Algorithm:

  1. Initialize variables left and right to 0 to represent the window boundaries.
  2. Initialize a hashmap letterCount to store the count of each letter in the window.
  3. Initialize variables minSubseq and minString to store the smallest subsequence found so far and its corresponding string.
  4. Iterate through the string s:
    • Increment the count of the current letter in the window.
    • Increment right to expand the window.
    • If the window size exceeds k, increment left and update the count of the letter accordingly.
    • Check if the current window contains at least repetition occurrences of the given letter:
      • If yes, update minSubseq and minString with the lexicographically smallest subsequence found.
  5. Return minSubseq.

Time Complexity:

The time complexity of this approach is O(n), where n is the length of the input string s.

Space Complexity:

The space complexity of this approach is O(1) since we are using a fixed-size hashmap to store letter counts.

Solution Code

class Solution {
    public String smallestSubsequence(String s, int k, char letter, int repetition) {
        int left = 0, right = 0, count = 0;
        int[] letterCount = new int[26];
        String minSubseq = "";
        String minString = "";
        
        for (right = 0; right < s.length(); right++) {
            if (s.charAt(right) == letter) {
                count++;
            }
            letterCount[s.charAt(right) - 'a']++;
            
            while (right - left + 1 > k) {
                if (s.charAt(left) == letter) {
                    count--;
                }
                letterCount[s.charAt(left) - 'a']--;
                left++;
            }
            
            if (count >= repetition && (minSubseq.equals("") || s.substring(left, right + 1).compareTo(minString) < 0)) {
                minSubseq = s.substring(left, right + 1);
                minString = minSubseq;
            }
        }
        
        return minSubseq;
    }
}

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