How do you solve Find the Maximum Length of Valid Subsequence I on LeetCode?

Find the Maximum Length of Valid Subsequence I (LeetCode #3201) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The validity of the subsequence depends solely on the parity of the numbers.

What is the brute force solution for Find the Maximum Length of Valid Subsequence I?

The brute force approach for Find the Maximum Length of Valid Subsequence I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible subsequences of the array to see if they are valid. This means we would generate every combination of elements and verify if they meet the validity condition based on their parities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Maximum Length of Valid Subsequence I?

Find the Maximum Length of Valid Subsequence I uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Counting elements based on conditions (like parity)., Dynamic programming concepts can sometimes apply, but this problem is more about counting..

What are the interview tips for Find the Maximum Length of Valid Subsequence I?

Always consider the properties of the elements (like parity) before diving into complex algorithms. Think about edge cases, such as arrays with all odd or all even numbers. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Find the Maximum Length of Valid Subsequence I?

Students often try to generate all subsequences without realizing the problem can be simplified by focusing on parity. Failing to account for both odd and even counts when determining the maximum length.

#3201

Find the Maximum Length of Valid Subsequence I

Medium

ArrayDynamic ProgrammingCounting elements based on conditions (like parity).Dynamic programming concepts can sometimes apply, but this problem is more about counting.

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution focuses on the parity of numbers instead of generating subsequences. We can count the number of odd and even numbers in the array and determine the longest valid subsequence based on their counts and possible valid patterns.

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Algorithm

3 steps

1Step 1: Initialize counters for odd and even numbers.
2Step 2: Iterate through the array and count the odd and even numbers.
3Step 3: The longest valid subsequence can be formed by using all odd or all even numbers, or by alternating them. Thus, the result is the maximum of the counts of odd and even numbers.

solution.py4 lines

1def maxLengthValidSubsequence(nums):
2    odd_count = sum(1 for num in nums if num % 2 != 0)
3    even_count = len(nums) - odd_count
4    return max(odd_count, even_count)

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array to count odd and even numbers. The space complexity is O(1) since we are using a fixed amount of extra space regardless of input size.

1The validity of the subsequence depends solely on the parity of the numbers.
2The longest valid subsequence can be formed by either all odd or all even numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.