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Find the Maximum Length of Valid Subsequence I

Medium

Google

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Topics:

Arrays

You are given an integer array nums.

A subsequence sub of nums with length x is called valid if it satisfies:

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

Return the length of the longest valid subsequence of nums.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

For example:

nums = [1,2,3,4] should return 4 because the longest valid subsequence is [1, 2, 3, 4].
nums = [1,2,1,1,2,1,2] should return 6 because the longest valid subsequence is [1, 2, 1, 2, 1, 2].
nums = [1,3] should return 2 because the longest valid subsequence is [1, 3].

Explain your approach and provide the time and space complexity.

Find the Maximum Length of Valid Subsequence I

Medium

Google

2 views

Topics:

Arrays

You are given an integer array nums.

A subsequence sub of nums with length x is called valid if it satisfies:

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

Return the length of the longest valid subsequence of nums.

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

For example:

nums = [1,2,3,4] should return 4 because the longest valid subsequence is [1, 2, 3, 4].
nums = [1,2,1,1,2,1,2] should return 6 because the longest valid subsequence is [1, 2, 1, 2, 1, 2].
nums = [1,3] should return 2 because the longest valid subsequence is [1, 3].

Explain your approach and provide the time and space complexity.

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:

What are the constraints on the values in the `nums` array and the `target`? Can they be negative, zero, or very large?
What should I return if no subsequence has a sum less than or equal to the target? Should I return 0, -1, or throw an exception?
Are duplicate numbers allowed in the `nums` array, and if so, do they affect the length calculation?
Is the order of elements in the subsequence important? In other words, does [1, 2] differ from [2, 1] if both are present in the original `nums` array?
Can the input array `nums` be empty or null?

Brute Force Solution

Approach

The brute force method for finding the longest valid subsequence means we will check every possible subsequence within the given sequence. We'll generate each subsequence and then test if it is 'valid' according to the problem's specific criteria.

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

Consider every possible group of elements you can create from the original sequence.
For each of these groups, examine if it forms a 'valid' subsequence based on the problem's rules (for example, ensuring a property such as elements being in ascending order or satisfying a specific mathematical relationship).
If a group is 'valid', remember its length.
After checking all possible groups, find the largest length among all the 'valid' subsequences you found. That's the answer.

Code Implementation

def find_maximum_length_of_valid_subsequence_brute_force(sequence):
    max_length = 0

    # Iterate through all possible subsequences
    for i in range(1 << len(sequence)):
        subsequence = []
        for j in range(len(sequence)):
            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_length = max(max_length, len(subsequence))

    return max_length

def is_valid_subsequence(subsequence):
    # In this dummy implementation, we consider every subsequence to be valid.
    # This should be replaced with your validation logic.
    return True

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach considers all possible subsequences. A sequence of length n has 2^n possible subsequences. For each subsequence, we need to validate if it's a 'valid' subsequence according to the problem constraints. This validation might take up to O(n) time in the worst-case scenario, where we have to iterate through each element of the subsequence to check the given property. Therefore, the overall time complexity is approximately O(2^n * n).

Space Complexity

O(1) – The brute force approach, as described, primarily involves iterating through possible subsequences and checking their validity. While generating each subsequence, it doesn't require significant auxiliary space beyond a few variables to track the current subsequence's length and the maximum length found so far. The space used to store these variables remains constant, irrespective of the input sequence's size, N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The core idea is to use a stack to maintain a subsequence that satisfies the given condition. We iterate through the input sequence, making decisions about whether to keep or discard elements to maximize the subsequence's length, ensuring validity at each step.

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

Imagine you have a box to store numbers, following certain rules.
Go through the numbers one by one.
For each number, check if putting it in the box breaks the rules.
If it breaks the rules, check the numbers currently in the box from most recent to oldest. Keep removing numbers from the box until it no longer breaks the rule when you add the current number.
Once putting the current number in the box doesn't break the rules, put it in.
After you've gone through all the numbers, count how many numbers are in the box. That's the length of the longest sequence you could make following the rules.

Code Implementation

def find_maximum_length_of_valid_subsequence(sequence):
    stack = []

    for element in sequence:
        # Remove elements from stack that violate condition.
        while stack and element < stack[-1]:

            stack.pop()

        # Add current element to the stack
        stack.append(element)

    # Stack size represents subsequence length
    return len(stack)

Big(O) Analysis

Time Complexity

O(n) – We iterate through the input sequence of size n once. For each element, we might perform multiple pop operations on the stack. However, each element can be pushed onto the stack at most once and therefore popped at most once. Thus, the total number of push and pop operations across all iterations is bounded by 2n. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The plain English explanation indicates using a 'box' to store numbers, which acts as a stack. In the worst-case scenario, where no numbers are removed from the 'box' during the iteration, the stack could potentially store all N numbers from the input sequence. Therefore, the auxiliary space required grows linearly with the input size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as no subsequence can be formed.
Input array contains only negative numbers and the target is negative.	Sort the array and greedily add elements until the sum exceeds the target.
Input array contains only positive numbers and the target is negative.	Return 0, as no subsequence can have a negative sum with positive elements.
Target is zero and the input array contains at least one zero.	The longest subsequence is the number of zeros in the array.
Target is very large (close to the maximum integer value) and the sum of all elements is less than or equal to it.	Return the length of the entire array.
Target is smaller than the smallest number in the array.	Return 0, as no element alone can satisfy the target constraint.
Input array contains very large numbers that could cause integer overflow when summing them.	Use long data type for sum accumulation to prevent potential overflow.
Array contains duplicate numbers and a single instance or multiple instances satisfies the target criteria. Example nums = [1,1,1,1,1], target = 1	Algorithm should count each duplicate if included without double-counting the index of elements.

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:

What are the constraints on the values in the `nums` array and the `target`? Can they be negative, zero, or very large?
What should I return if no subsequence has a sum less than or equal to the target? Should I return 0, -1, or throw an exception?
Are duplicate numbers allowed in the `nums` array, and if so, do they affect the length calculation?
Is the order of elements in the subsequence important? In other words, does [1, 2] differ from [2, 1] if both are present in the original `nums` array?
Can the input array `nums` be empty or null?

Brute Force Solution

Approach

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

Consider every possible group of elements you can create from the original sequence.
For each of these groups, examine if it forms a 'valid' subsequence based on the problem's rules (for example, ensuring a property such as elements being in ascending order or satisfying a specific mathematical relationship).
If a group is 'valid', remember its length.
After checking all possible groups, find the largest length among all the 'valid' subsequences you found. That's the answer.

Code Implementation

def find_maximum_length_of_valid_subsequence_brute_force(sequence):
    max_length = 0

    # Iterate through all possible subsequences
    for i in range(1 << len(sequence)):
        subsequence = []
        for j in range(len(sequence)):
            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_length = max(max_length, len(subsequence))

    return max_length

def is_valid_subsequence(subsequence):
    # In this dummy implementation, we consider every subsequence to be valid.
    # This should be replaced with your validation logic.
    return True

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you have a box to store numbers, following certain rules.
Go through the numbers one by one.
For each number, check if putting it in the box breaks the rules.
If it breaks the rules, check the numbers currently in the box from most recent to oldest. Keep removing numbers from the box until it no longer breaks the rule when you add the current number.
Once putting the current number in the box doesn't break the rules, put it in.
After you've gone through all the numbers, count how many numbers are in the box. That's the length of the longest sequence you could make following the rules.

Code Implementation

def find_maximum_length_of_valid_subsequence(sequence):
    stack = []

    for element in sequence:
        # Remove elements from stack that violate condition.
        while stack and element < stack[-1]:

            stack.pop()

        # Add current element to the stack
        stack.append(element)

    # Stack size represents subsequence length
    return len(stack)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as no subsequence can be formed.
Input array contains only negative numbers and the target is negative.	Sort the array and greedily add elements until the sum exceeds the target.
Input array contains only positive numbers and the target is negative.	Return 0, as no subsequence can have a negative sum with positive elements.
Target is zero and the input array contains at least one zero.	The longest subsequence is the number of zeros in the array.
Target is very large (close to the maximum integer value) and the sum of all elements is less than or equal to it.	Return the length of the entire array.
Target is smaller than the smallest number in the array.	Return 0, as no element alone can satisfy the target constraint.
Input array contains very large numbers that could cause integer overflow when summing them.	Use long data type for sum accumulation to prevent potential overflow.
Array contains duplicate numbers and a single instance or multiple instances satisfies the target criteria. Example nums = [1,1,1,1,1], target = 1	Algorithm should count each duplicate if included without double-counting the index of elements.