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

Find the Maximum Length of Valid Subsequence II (LeetCode #3202) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(k) space complexity. Key insight: The validity condition relies on the remainders when divided by k, which allows us to group numbers effectively.

What is the time complexity of Find the Maximum Length of Valid Subsequence II?

The optimal solution for Find the Maximum Length of Valid Subsequence II runs in O(n) time and uses O(k) space. The time complexity is O(n) because we iterate through the array once, and the space complexity is O(k) due to the dp array.

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

The brute force approach for Find the Maximum Length of Valid Subsequence II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences of the array and checking each one to see if it meets the validity condition. This is straightforward but inefficient. 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 II?

Find the Maximum Length of Valid Subsequence II uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

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

Clearly explain your thought process and how you derive the optimal solution from the brute-force approach. Practice writing clean and efficient code, as interviewers often look for both correctness and efficiency. Be prepared to discuss the time and space complexities of your solutions.

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

Failing to consider all possible remainders when updating the dp array. Not correctly implementing the condition check for subsequences.

#3202

Find the Maximum Length of Valid Subsequence II

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(k)

The optimal approach uses dynamic programming to store the maximum length of valid subsequences based on the remainder when the last element is divided by k. This reduces the problem size significantly.

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Algorithm

3 steps

1Step 1: Initialize a dp array of size k to store the maximum length of subsequences for each possible remainder.
2Step 2: Iterate through each number in nums and update the dp array based on the current number's contribution to valid subsequences.
3Step 3: For each number, calculate its remainder and update the dp array accordingly.

solution.py12 lines

1# Full working Python code
2
3def maxLengthValidSubsequence(nums, k):
4    dp = [0] * k
5    for num in nums:
6        rem = num % k
7        new_dp = dp[:]
8        for i in range(k):
9            if (i + rem) % k == (rem + rem) % k:
10                new_dp[(i + rem) % k] = max(new_dp[(i + rem) % k], dp[i] + 1)
11        dp = new_dp
12    return max(dp)

ℹ

Complexity note: The time complexity is O(n) because we iterate through the array once, and the space complexity is O(k) due to the dp array.

1The validity condition relies on the remainders when divided by k, which allows us to group numbers effectively.
2Dynamic programming can significantly reduce the complexity by storing intermediate results.

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