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LeetCode 3176: Find the Maximum Length of a Good Subsequence I Solution

Master LeetCode problem 3176 (Find the Maximum Length of a Good Subsequence I), 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.

3176. Find the Maximum Length of a Good Subsequence I

Medium
ArrayHash TableDynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use a greedy approach where we iterate through the input array nums and construct the subsequence by selecting the maximum element at each step while ensuring that the conditions are met. We can keep track of the count of consecutive different elements encountered so far and use this information to limit the selection of elements in the subsequence.

Algorithm

  1. Sort the input array nums in non-decreasing order.
  2. Initialize a variable count to keep track of the count of consecutive different elements encountered.
  3. Iterate through the sorted array and greedily select each element to form the subsequence, ensuring that the count of consecutive different elements does not exceed k.
  4. Return the length of the subsequence.

Time Complexity

The time complexity of this algorithm is O(n log n) due to the sorting of the input array.

Space Complexity

The space complexity of this algorithm is O(n) for storing the sorted array.

Solution Code

import java.util.Arrays;

class Solution {
    public int maxGoodSubsequence(int[] nums, int k) {
        Arrays.sort(nums);
        int count = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i == 0 || nums[i] != nums[i - 1] || count < k) {
                count++;
            }
        }
        return count;
    }
}

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What is the time complexity of LeetCode 3176 (Find the Maximum Length of a Good Subsequence I)?

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