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Find the Maximum Length of Valid Subsequence II

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ArraysDynamic Programming

You are given an integer array nums and a positive integer k. A subsequence sub of nums with length x is called valid if it satisfies:

  • (sub[0] + sub[1]) % k == (sub[1] + sub[2]) % k == ... == (sub[x - 2] + sub[x - 1]) % k.

Return the length of the longest valid subsequence of nums.

For example:

nums = [1,2,3,4,5], k = 2

The longest valid subsequence is [1, 2, 3, 4, 5]. The function should return 5.

Another example:

nums = [1,4,2,3,1,4], k = 3

The longest valid subsequence is [1, 4, 1, 4]. The function should return 4.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. What are the constraints on the input array's size and the range of values within it?
  2. Can the input array contain duplicate numbers, and if so, how should they be handled?
  3. If there are multiple valid subsequences of the maximum length, is any one acceptable, or is a specific subsequence required?
  4. Is an empty array a valid input, and if so, what should be the returned length?
  5. Is the term 'valid subsequence' based on a specific property (e.g. sorted order, difference between elements, divisibility), and if so, could you elaborate?

Brute Force Solution

Approach

The brute force strategy for this problem involves examining every single possible subsequence to find the longest valid one. We generate all possible combinations and then check if each one meets the problem's criteria for being valid. Finally, we compare the lengths of all the valid subsequences and select the maximum.

Here's how the algorithm would work step-by-step:

  1. Consider all possible combinations of the given sequence, from just a single element to the entire sequence itself.
  2. For each of these combinations, verify if it's a valid subsequence according to the specific rules outlined in the problem description (e.g., checking if elements meet a certain ratio or condition).
  3. If a subsequence is valid, remember its length.
  4. After checking all combinations, identify the longest length among all the valid subsequences you've found.
  5. The longest length identified in the previous step is the result: the maximum length of a valid subsequence.

Code Implementation

def find_max_length_valid_subsequence_brute_force(sequence):
    max_subsequence_length = 0

    # Iterate through all possible subsequences
    for i in range(1 << len(sequence)):
        subsequence = []
        for j in range(len(sequence)):
            # Check if the j-th element is included in the current subsequence
            if (i >> j) & 1:
                subsequence.append(sequence[j])

        # Check if the current subsequence is valid.
        if is_valid_subsequence(subsequence):

            # Update the maximum length if the current subsequence is longer.
            max_subsequence_length = max(max_subsequence_length, len(subsequence))

    return max_subsequence_length

def is_valid_subsequence(subsequence):
    if not subsequence:
        return True
    
    for i in range(len(subsequence) - 1):
        if subsequence[i+1] <= subsequence[i]:
            return False
    return True

Big(O) Analysis

Time Complexity
O(2^n * n)The brute force approach involves generating all possible subsequences of the input sequence, which takes O(2^n) time because each element can either be included or excluded from a subsequence. For each subsequence generated, we then need to validate it according to the problem's specific rules. The validation step iterates through the subsequence, which, in the worst case, can have length n. Therefore, the overall time complexity is O(2^n * n).
Space Complexity
O(1)The brute force solution, as described, generates all possible subsequences. However, it doesn't explicitly mention storing all subsequences simultaneously. It keeps track of the longest valid subsequence length found so far. The only auxiliary space used is for storing the current subsequence length, and the maximum length found so far, both of which are constant. Thus, the auxiliary space used is independent of the input size N, resulting in constant space complexity.

Optimal Solution

Approach

The key is to use a technique that efficiently skips over subsequences that are guaranteed to be invalid. We keep track of the best possible length we have seen so far to help us decide whether or not to explore a specific subsequence. This technique helps significantly reduce the search space.

Here's how the algorithm would work step-by-step:

  1. First, sort the numbers in the input sequence. This will help us easily identify valid subsequences.
  2. Keep track of the longest valid subsequence found so far.
  3. Now, consider each number one by one. For each number, think of it as potentially the starting point of a new valid subsequence.
  4. For each starting number, extend the subsequence by looking at the remaining numbers after it.
  5. A number can be added to the subsequence only if it meets the condition where each number is greater than or equal to twice the previous number.
  6. As you build each potential subsequence, compare its length to the longest valid subsequence you've seen so far and update if necessary.
  7. Crucially, if at any point while building a subsequence, the remaining numbers are not enough to potentially exceed the current longest valid length, then abandon that subsequence. This early cutoff avoids unnecessary computations and significantly improves efficiency.
  8. Repeat this process for all possible starting points, and in the end, the longest valid subsequence found will be the answer.

Code Implementation

def find_maximum_length_of_valid_subsequence(sequence):
    sequence.sort()
    longest_subsequence_length = 0

    for starting_index in range(len(sequence)): 
        current_subsequence_length = 0
        last_element = -1
        
        def extend_subsequence(current_index):
            nonlocal current_subsequence_length
            nonlocal longest_subsequence_length
            nonlocal last_element

            # Early cutoff: Not enough elements remaining.
            if current_subsequence_length + (len(sequence) - current_index) < longest_subsequence_length:
                return
            
            if current_index == len(sequence):
                longest_subsequence_length = max(longest_subsequence_length, current_subsequence_length)
                return

            # Check if we can extend the subsequence.
            if sequence[current_index] >= 2 * last_element:
                current_length_before = current_subsequence_length
                last_element = sequence[current_index]
                current_subsequence_length += 1
                extend_subsequence(current_index + 1)
                current_subsequence_length = current_length_before
                last_element = sequence[starting_index] if current_subsequence_length > 0 else -1

            # Explore the possibility of skipping this element.
            extend_subsequence(current_index + 1)

        #Consider each element as the potential start.
        current_subsequence_length = 1
        last_element = sequence[starting_index]
        extend_subsequence(starting_index + 1)

    return longest_subsequence_length

Big(O) Analysis

Time Complexity
O(n log n)Sorting the input array of n elements takes O(n log n) time. Then, for each element, we potentially iterate through the remaining elements to build a valid subsequence. The number of such iterations is pruned based on the current maximum length found so far. In the worst case, for each of the n elements, we may still need to traverse a significant portion of the remaining array elements, though fewer than a nested loop due to the subsequence length check. Therefore, the time complexity is dominated by sorting, with the subsequence building having an average cost faster than a completely nested loop, so O(n log n).
Space Complexity
O(1)The algorithm sorts the input sequence in place, modifying the original input array. Beyond that, it only stores a few variables like the current longest subsequence length and indices for iteration. No auxiliary data structures that scale with the input size N (number of elements in the input sequence) are used. Therefore, the space complexity is constant.

Edge Cases

CaseHow to Handle
Empty input arrayReturn 0 immediately as an empty array cannot contain a valid subsequence.
Input array with only one elementReturn 0 because a subsequence requires at least two elements, thus one element will always be invalid.
Array with all identical values.Iterate through the array, for each element count the duplicates of the elements that are exactly twice the current element and those that are exactly half and choose the subsequence with maximum length.
Array contains negative numbers.The solution should handle negative numbers correctly by considering both multiplication and division by 2 for subsequence verification, ensure no integer overflow occurs during multiplication.
Array contains zero.Handle zero as special case; any zero multiplied by 2 is zero again so the subsequence should not include more than 1 zero, or division by zero errors will occur.
Integer overflow during multiplication by 2Use long data type or perform checks to prevent overflow during multiplication, potentially capping the maximum value before multiplication.
Input array with very large number of elements (performance)Optimize the solution to reduce time complexity, potentially using memoization or dynamic programming depending on the specific algorithm used to determine subsequence validity.
No valid subsequence exists in the inputThe algorithm should return 0 when no valid subsequence is found, such as if all numbers are prime or are unrelated by factors of 2.