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Find the Maximum Length of a Good Subsequence I

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Dynamic 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 <= 500
1 <= nums[i] <= 10⁹
0 <= k <= min(nums.length, 25)

Try coding it in LeetCode

Find the Maximum Length of a Good Subsequence I

Medium

Asked by:

2 views

Topics:

Dynamic 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 <= 500
1 <= nums[i] <= 10⁹
0 <= k <= min(nums.length, 25)

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 input array `nums` and the value of `k`?
Can the integers in the `nums` array be negative, zero, or only positive?
If no good subsequence exists, what value should I return?
Are there any specific memory constraints I should be aware of?
If multiple good subsequences exist with the maximum length, is any one of them acceptable?

Brute Force Solution

Approach

The brute force approach involves checking every possible subsequence of the given sequence. We'll generate each subsequence, verify if it meets the 'good' condition, and keep track of the longest one encountered. By checking every possibility, we guarantee finding the maximum length good subsequence.

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

Start by considering all the shortest subsequences, each containing only a single element from the original sequence.
Check if each of these single-element subsequences is 'good' according to the problem's definition.
Then, move on to all possible subsequences with two elements. For example, take the first two elements, then the first and third, then the first and fourth, and so on.
Again, check if each two-element subsequence is 'good'.
Continue this process, building subsequences of increasing length (three elements, four elements, and so on) until we have considered all possible subsequences, including the entire original sequence itself.
Each time we find a 'good' subsequence, compare its length to the length of the longest 'good' subsequence we've found so far. If the new one is longer, remember it.
After we have checked all possible subsequences, the longest 'good' subsequence we remembered is the answer.

Code Implementation

def find_maximum_length_of_good_subsequence_brute_force(sequence):
    maximum_length_of_good_subsequence = 0
    number_of_elements = len(sequence)

    # Iterate through all possible subsequences
    for i in range(1 << number_of_elements):
        current_subsequence = []
        for j in range(number_of_elements):
            # Check if j-th bit is set in the current combination
            if (i >> j) & 1:
                current_subsequence.append(sequence[j])

        # Check if the current subsequence is "good"
        is_good = True
        if len(current_subsequence) > 1:
            for k in range(len(current_subsequence) - 1):
                if current_subsequence[k] >= current_subsequence[k+1]:
                    is_good = False
                    break

        # Update the maximum length if the current subsequence is "good"
        if is_good:
            length_of_current_subsequence = len(current_subsequence)
            maximum_length_of_good_subsequence = max(
                maximum_length_of_good_subsequence,
                length_of_current_subsequence
            )

    return maximum_length_of_good_subsequence

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves generating all possible subsequences of the input sequence. For a sequence of length n, there are 2^n possible subsequences (each element can either be included or excluded in a subsequence). The algorithm checks each of these subsequences to see if it's 'good'. Therefore, the algorithm's time complexity is directly proportional to the number of subsequences it generates and checks which is 2^n. Thus, the time complexity is O(2^n).

Space Complexity

O(1) – The described brute force approach iterates through all possible subsequences, but the provided plain English explanation does not explicitly mention the creation of any auxiliary data structures to store these subsequences. The algorithm seems to be directly comparing subsequences without storing them. Therefore, the auxiliary space complexity is constant, as it only requires a few variables to keep track of the longest good subsequence found so far, irrespective of the input sequence length N.

Optimal Solution

Approach

The goal is to find the longest possible sequence of numbers from the input where no two adjacent numbers in the sequence are the same. We can achieve this by selectively including numbers from the original input to build our 'good' sequence. We don't need to check all possible sequences.

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

Go through the numbers one by one, from beginning to end.
If the current number is different from the last number we added to our sequence, then add it to the sequence.
If the current number is the same as the last number we added, then we skip it and move to the next number.
Keep doing this until you have checked all the numbers in the original input.
The length of the sequence you've built is the maximum length of a 'good' subsequence.

Code Implementation

def find_maximum_length_of_good_subsequence(input_sequence):
    good_subsequence = []

    for element in input_sequence:
        #The subsequence is empty, so add the first element.
        if not good_subsequence:
            good_subsequence.append(element)
            continue

        #Check if adding the current element maintains the good property.
        if element > good_subsequence[-1]:

            good_subsequence.append(element)

        #Otherwise skip the element and move to the next.

    return len(good_subsequence)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input list of numbers once. For each number, it performs a constant-time comparison to the last element added to the subsequence. Since the number of comparisons and additions is directly proportional to the size of the input list (n), the time complexity is O(n).

Space Complexity

O(1) – The algorithm maintains only a single variable to store the last number added to the sequence. No other data structures that scale with the input size are used. Therefore, the auxiliary space required is constant, irrespective of the input size N. Hence, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array (nums is null or has length 0)	Return 0, as there are no elements to form a subsequence.
k is 0	If k is zero, return the count of zero elements if any exist, otherwise return 0.
Array contains only zeros	If k is not zero return the length of the array, if k is zero return the array length.
Array contains only one element	If the element is divisible by k return 1, otherwise return 0.
Array contains negative numbers	The dynamic programming solution should handle negative numbers correctly by considering negative remainders.
All numbers have the same remainder when divided by k	If the sum of the numbers is divisible by k, the length of the array is the answer; otherwise, the answer is 0 if the array is empty, and otherwise the array length minus 1 if the array is not empty.
Large array size leading to potential memory issues	Ensure dynamic programming arrays are appropriately sized and optimized for space efficiency, potentially using modulo operations to reduce the range of remainders.
k is negative	Take the absolute value of k as the remainder calculation works the same for positive and negative k after that transformation.

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 input array `nums` and the value of `k`?
Can the integers in the `nums` array be negative, zero, or only positive?
If no good subsequence exists, what value should I return?
Are there any specific memory constraints I should be aware of?
If multiple good subsequences exist with the maximum length, is any one of them acceptable?

Brute Force Solution

Approach

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

Start by considering all the shortest subsequences, each containing only a single element from the original sequence.
Check if each of these single-element subsequences is 'good' according to the problem's definition.
Then, move on to all possible subsequences with two elements. For example, take the first two elements, then the first and third, then the first and fourth, and so on.
Again, check if each two-element subsequence is 'good'.
Continue this process, building subsequences of increasing length (three elements, four elements, and so on) until we have considered all possible subsequences, including the entire original sequence itself.
Each time we find a 'good' subsequence, compare its length to the length of the longest 'good' subsequence we've found so far. If the new one is longer, remember it.
After we have checked all possible subsequences, the longest 'good' subsequence we remembered is the answer.

Code Implementation

def find_maximum_length_of_good_subsequence_brute_force(sequence):
    maximum_length_of_good_subsequence = 0
    number_of_elements = len(sequence)

    # Iterate through all possible subsequences
    for i in range(1 << number_of_elements):
        current_subsequence = []
        for j in range(number_of_elements):
            # Check if j-th bit is set in the current combination
            if (i >> j) & 1:
                current_subsequence.append(sequence[j])

        # Check if the current subsequence is "good"
        is_good = True
        if len(current_subsequence) > 1:
            for k in range(len(current_subsequence) - 1):
                if current_subsequence[k] >= current_subsequence[k+1]:
                    is_good = False
                    break

        # Update the maximum length if the current subsequence is "good"
        if is_good:
            length_of_current_subsequence = len(current_subsequence)
            maximum_length_of_good_subsequence = max(
                maximum_length_of_good_subsequence,
                length_of_current_subsequence
            )

    return maximum_length_of_good_subsequence

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Go through the numbers one by one, from beginning to end.
If the current number is different from the last number we added to our sequence, then add it to the sequence.
If the current number is the same as the last number we added, then we skip it and move to the next number.
Keep doing this until you have checked all the numbers in the original input.
The length of the sequence you've built is the maximum length of a 'good' subsequence.

Code Implementation

def find_maximum_length_of_good_subsequence(input_sequence):
    good_subsequence = []

    for element in input_sequence:
        #The subsequence is empty, so add the first element.
        if not good_subsequence:
            good_subsequence.append(element)
            continue

        #Check if adding the current element maintains the good property.
        if element > good_subsequence[-1]:

            good_subsequence.append(element)

        #Otherwise skip the element and move to the next.

    return len(good_subsequence)

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 there are no elements to form a subsequence.
k is 0	If k is zero, return the count of zero elements if any exist, otherwise return 0.
Array contains only zeros	If k is not zero return the length of the array, if k is zero return the array length.
Array contains only one element	If the element is divisible by k return 1, otherwise return 0.
Array contains negative numbers	The dynamic programming solution should handle negative numbers correctly by considering negative remainders.
All numbers have the same remainder when divided by k	If the sum of the numbers is divisible by k, the length of the array is the answer; otherwise, the answer is 0 if the array is empty, and otherwise the array length minus 1 if the array is not empty.
Large array size leading to potential memory issues	Ensure dynamic programming arrays are appropriately sized and optimized for space efficiency, potentially using modulo operations to reduce the range of remainders.
k is negative	Take the absolute value of k as the remainder calculation works the same for positive and negative k after that transformation.