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Find the Maximum Length of a Good Subsequence II

Hard

Asked by:

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Topics:

ArraysDynamic Programming

You are given an integer array nums and a non-negative integer k. A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1].

Return the maximum possible length of a good subsequence of nums.

Example 1:

Input: nums = [1,2,1,1,3], k = 2

Output: 4

Explanation:

The maximum length subsequence is [1,2,1,1,3].

Example 2:

Input: nums = [1,2,3,4,5,1], k = 0

Output: 2

Explanation:

The maximum length subsequence is [1,2,3,4,5,1].

Constraints:

1 <= nums.length <= 5 * 10³
1 <= nums[i] <= 10⁹
0 <= k <= min(50, nums.length)

Try coding it in LeetCode

Find the Maximum Length of a Good Subsequence II

Hard

Asked by:

3 views

Topics:

ArraysDynamic Programming

Return the maximum possible length of a good subsequence of nums.

Example 1:

Input: nums = [1,2,1,1,3], k = 2

Output: 4

Explanation:

The maximum length subsequence is [1,2,1,1,3].

Example 2:

Input: nums = [1,2,3,4,5,1], k = 0

Output: 2

Explanation:

The maximum length subsequence is [1,2,3,4,5,1].

Constraints:

1 <= nums.length <= 5 * 10³
1 <= nums[i] <= 10⁹
0 <= k <= min(50, nums.length)

Try coding it in LeetCode

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 size of the `nums` array, the range of values in `nums`, the value of `k`, and the value of `target`?
Can `k` or `target` be zero or negative?
If no good subsequence exists, what value should I return?
If there are multiple good subsequences with the same maximum length, is any one of them acceptable, or should I return something specific?
Are duplicate numbers allowed in `nums`, and if so, do they affect the definition of a good subsequence or its length?

Brute Force Solution

Approach

The brute force approach to finding the longest 'good' subsequence means we examine every single possible subsequence. We then check if each subsequence we find is a 'good' one based on the given rules. Finally, we track the length of each good subsequence, and pick the longest one we've encountered.

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

First, consider all possible groups of numbers you can make from the original set of numbers. This means starting with groups of just one number.
Then, consider groups of two numbers, then three numbers, and so on, until you have considered groups containing all the numbers from the original set.
For each of these groups, determine if it qualifies as a 'good' subsequence. This might involve checking if the numbers in the group meet specific criteria or relationships defined by the problem.
If a group is 'good', record its length (the number of items in the group).
After checking all possible groups and recording the lengths of the 'good' ones, find the largest length you recorded. That's the length of the longest good subsequence.

Code Implementation

def find_maximum_length_of_a_good_subsequence_brute_force(sequence):
    maximum_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 subsequence is 'good'
        if is_good_subsequence(subsequence):
            current_length = len(subsequence)

            # Update the maximum length if necessary
            if current_length > maximum_length:

                # Keep track of the longest good subsequence found so far.
                maximum_length = current_length

    return maximum_length

def is_good_subsequence(subsequence):
    if not subsequence:
        return True
    
    #This is a dummy implementation and the good subsequence needs to be defined
    for i in range(len(subsequence) - 1):
        if subsequence[i] > subsequence[i+1]:
            return False

    return True

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves generating all possible subsequences of the input array. For an array of size n, there are 2^n possible subsequences (each element can either be included or excluded). For each subsequence, we need to perform checks to determine if it's a 'good' subsequence, which takes O(n) time in the worst case where we might need to examine all the elements of the subsequence. Because the number of subsequence increases exponentially, this becomes the dominant factor. Thus the overall time complexity is O(n*2^n). Simplifying and focusing on the dominant term, we can approximate the time complexity as O(2^n).

Space Complexity

O(2^N) – The algorithm generates all possible subsequences of the input set. Creating each subsequence requires storing its elements in memory. In the worst case, there are 2^N possible subsequences for an input of size N, where N is the number of elements in the original set. Therefore, the auxiliary space used to store these subsequences grows exponentially with the input size. Checking each subsequence also requires some constant space, but the dominant term is the space to store the subsequences themselves.

Optimal Solution

Approach

The goal is to find the longest possible subsequence in a given sequence that meets specific criteria, efficiently. Instead of checking every possible subsequence which would take too long, we'll build up the solution step by step, remembering the best results we've seen so far.

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

Start by sorting the sequence. This makes it easier to find the elements that form a good subsequence.
Create a way to remember the best (longest) subsequence we can make ending at each element in the sequence.
Go through the sequence, element by element.
For each element, check if it can extend any of the good subsequences we already found. This means checking if it fits the criteria alongside the last element of those subsequences.
If it can extend a subsequence, we now have a new, potentially better, subsequence. Update our records of the best subsequences accordingly.
We can avoid checking every single existing subsequence. We only need to keep track of the shortest possible length of subsequence for a given ending value. If the current element extends a subsequence, and the length of that new subsequence is shorter than the current shortest-possible-subsequence with that same ending value, then we can use it to update the shortest length.
After checking all elements, the length of the best subsequence we remembered is the answer.

Code Implementation

def find_maximum_length_of_good_subsequence(numbers, factor):
    smallest_last_numbers = {}

    for number in numbers:
        extendable = False

        for length, last_number in smallest_last_numbers.items():
            # Check if the current number extends the sequence.
            if number >= last_number * factor:
                extendable = True
                new_length = length + 1

                # Update if this is the smallest last number for this length.
                if new_length not in smallest_last_numbers or number < smallest_last_numbers[new_length]:
                    smallest_last_numbers[new_length] = number

        # Start a new subsequence if it can't extend existing ones.
        if not extendable:
            if 1 not in smallest_last_numbers or number < smallest_last_numbers[1]:
                smallest_last_numbers[1] = number

    # Find the maximum length among the created subsequences.
    if not smallest_last_numbers:
        return 0

    return max(smallest_last_numbers.keys())

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm first sorts the input array of size n, which takes O(n log n) time. Then, it iterates through each of the n elements. Inside the loop, for each element, the algorithm potentially needs to find a suitable subsequence to extend. To optimize this search, the explanation implies an update operation which is logarithmic. Therefore, the dominant time complexity comes from iterating through n elements and performing a logarithmic operation at each, resulting in an O(n log n) time complexity inside the main loop. Since the sorting step is also O(n log n), the overall time complexity is O(n log n).

Space Complexity

O(N) – The solution uses a data structure (as described in step 2 and implicitly in steps 5 and 6) to remember the best subsequence ending at each element. This implies storing some value (e.g., the shortest length) for each of the N elements in the input sequence. Therefore, we have an auxiliary array (or hash map) whose size scales linearly with the input size N. Consequently, the space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0 immediately as no subsequence can be formed.
k is 0	Return 0 immediately, as no number is divisible by zero, so no good subsequence exists.
target is 0	If there are zero values in nums which are divisible by k, the answer is the number of zero values; otherwise return 0.
nums contains only negative numbers and the target is positive	Return 0, since the sum of negative numbers cannot equal a positive target.
No element in nums is divisible by k	Return 0 because no 'good' subsequence can be formed.
Very large nums array, potentially leading to memory issues with dynamic programming.	Optimize dynamic programming space by only storing necessary states, or consider an iterative approach if memory becomes a constraint.
Target is a very large number, potentially leading to integer overflow during summation.	Use a data type with a larger range (e.g., long) to store intermediate sums, or check for overflows before performing the addition.
Multiple 'good' subsequences exist, but they have different lengths.	The solution should find the maximum length among all valid 'good' subsequences.

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 size of the `nums` array, the range of values in `nums`, the value of `k`, and the value of `target`?
Can `k` or `target` be zero or negative?
If no good subsequence exists, what value should I return?
If there are multiple good subsequences with the same maximum length, is any one of them acceptable, or should I return something specific?
Are duplicate numbers allowed in `nums`, and if so, do they affect the definition of a good subsequence or its length?

Brute Force Solution

Approach

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

First, consider all possible groups of numbers you can make from the original set of numbers. This means starting with groups of just one number.
Then, consider groups of two numbers, then three numbers, and so on, until you have considered groups containing all the numbers from the original set.
For each of these groups, determine if it qualifies as a 'good' subsequence. This might involve checking if the numbers in the group meet specific criteria or relationships defined by the problem.
If a group is 'good', record its length (the number of items in the group).
After checking all possible groups and recording the lengths of the 'good' ones, find the largest length you recorded. That's the length of the longest good subsequence.

Code Implementation

def find_maximum_length_of_a_good_subsequence_brute_force(sequence):
    maximum_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 subsequence is 'good'
        if is_good_subsequence(subsequence):
            current_length = len(subsequence)

            # Update the maximum length if necessary
            if current_length > maximum_length:

                # Keep track of the longest good subsequence found so far.
                maximum_length = current_length

    return maximum_length

def is_good_subsequence(subsequence):
    if not subsequence:
        return True
    
    #This is a dummy implementation and the good subsequence needs to be defined
    for i in range(len(subsequence) - 1):
        if subsequence[i] > subsequence[i+1]:
            return False

    return True

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by sorting the sequence. This makes it easier to find the elements that form a good subsequence.
Create a way to remember the best (longest) subsequence we can make ending at each element in the sequence.
Go through the sequence, element by element.
For each element, check if it can extend any of the good subsequences we already found. This means checking if it fits the criteria alongside the last element of those subsequences.
If it can extend a subsequence, we now have a new, potentially better, subsequence. Update our records of the best subsequences accordingly.
We can avoid checking every single existing subsequence. We only need to keep track of the shortest possible length of subsequence for a given ending value. If the current element extends a subsequence, and the length of that new subsequence is shorter than the current shortest-possible-subsequence with that same ending value, then we can use it to update the shortest length.
After checking all elements, the length of the best subsequence we remembered is the answer.

Code Implementation

def find_maximum_length_of_good_subsequence(numbers, factor):
    smallest_last_numbers = {}

    for number in numbers:
        extendable = False

        for length, last_number in smallest_last_numbers.items():
            # Check if the current number extends the sequence.
            if number >= last_number * factor:
                extendable = True
                new_length = length + 1

                # Update if this is the smallest last number for this length.
                if new_length not in smallest_last_numbers or number < smallest_last_numbers[new_length]:
                    smallest_last_numbers[new_length] = number

        # Start a new subsequence if it can't extend existing ones.
        if not extendable:
            if 1 not in smallest_last_numbers or number < smallest_last_numbers[1]:
                smallest_last_numbers[1] = number

    # Find the maximum length among the created subsequences.
    if not smallest_last_numbers:
        return 0

    return max(smallest_last_numbers.keys())

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 immediately as no subsequence can be formed.
k is 0	Return 0 immediately, as no number is divisible by zero, so no good subsequence exists.
target is 0	If there are zero values in nums which are divisible by k, the answer is the number of zero values; otherwise return 0.
nums contains only negative numbers and the target is positive	Return 0, since the sum of negative numbers cannot equal a positive target.
No element in nums is divisible by k	Return 0 because no 'good' subsequence can be formed.
Very large nums array, potentially leading to memory issues with dynamic programming.	Optimize dynamic programming space by only storing necessary states, or consider an iterative approach if memory becomes a constraint.
Target is a very large number, potentially leading to integer overflow during summation.	Use a data type with a larger range (e.g., long) to store intermediate sums, or check for overflows before performing the addition.
Multiple 'good' subsequences exist, but they have different lengths.	The solution should find the maximum length among all valid 'good' subsequences.