LeetCode 3201: Find the Maximum Length of Valid Subsequence I Solution

Master LeetCode problem 3201 (Find the Maximum Length of Valid 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.

3201. Find the Maximum Length of Valid Subsequence I

Medium

Array Dynamic Programming

Problem Explanation

Explanation

To find the maximum length of a valid subsequence in the given array, we can iterate through the array and keep track of the maximum length of a valid subsequence ending at each index. We can use dynamic programming to solve this problem efficiently.

Initialize an array dp of size nums.length to store the maximum length of a valid subsequence ending at each index.
Initialize all elements of dp to 2, as a subsequence of length 2 is always valid.
Iterate through the array starting from the second element.
For each element at index i, check if the difference between nums[i] and nums[i-1] is even.
If it is even, update dp[i] = dp[i-1] + 1, as we can extend the valid subsequence ending at index i.
Otherwise, leave dp[i] as is, as we cannot extend the valid subsequence ending at index i.
Finally, return the maximum value in the dp array, which represents the length of the longest valid subsequence.

Time complexity: O(n) Space complexity: O(n)

Solution Code

class Solution {
    public int maxSubsequenceLength(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n];
        Arrays.fill(dp, 2);
        
        for (int i = 2; i < n; i++) {
            if ((nums[i] - nums[i-1]) % 2 == 0) {
                dp[i] = dp[i-1] + 1;
            }
        }
        
        int maxLen = Arrays.stream(dp).max().getAsInt();
        return maxLen;
    }
}

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