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Non-decreasing Subsequences

Medium

Amazon

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Topics:

RecursionArrays

Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.

For example:

nums = [4,6,7,7] Output: [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]

nums = [4,4,3,2,1] Output: [[4,4]]

Clarify the requirements and constraints of the problem. Consider edge cases such as:

What should happen if the input array is empty?
What should happen if all elements are decreasing?
What should happen if the input array has duplicate numbers?

Write a function that efficiently finds all non-decreasing subsequences with at least two elements.

Non-decreasing Subsequences

Medium

Amazon

3 views

Topics:

RecursionArrays

Given an integer array nums, return all the different possible non-decreasing subsequences of the given array with at least two elements. You may return the answer in any order.

For example:

nums = [4,6,7,7] Output: [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]

nums = [4,4,3,2,1] Output: [[4,4]]

Clarify the requirements and constraints of the problem. Consider edge cases such as:

What should happen if the input array is empty?
What should happen if all elements are decreasing?
What should happen if the input array has duplicate numbers?

Write a function that efficiently finds all non-decreasing subsequences with at least two 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:

Can the input array contain duplicate numbers, and if so, are duplicates allowed within a non-decreasing subsequence?
What is the expected output if the input array is empty?
What is the data type of the numbers in the input array, and are there any constraints on the range of values?
Should the returned subsequences be sorted in any specific order?
Is the input array guaranteed to have at least one element?

Brute Force Solution

Approach

The brute force approach to finding non-decreasing subsequences involves examining every possible subsequence within the given sequence. It's like trying out every single combination, no matter how big or small, to see if it fits the requirements. We then check if it meets the non-decreasing criteria.

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

Start by considering a subsequence containing only the first element of the original sequence.
Then, consider all subsequences containing the first two elements, then the first three, and so on, until you've considered subsequences starting with the first element and using all other elements.
Next, repeat the process, but start with the second element as the beginning of the subsequence, and consider all combinations of elements after it.
Continue doing this, shifting the starting element one by one to the right, until you have considered starting with every possible element of the original sequence.
For each subsequence you consider, check if its elements are in non-decreasing order (each element is greater than or equal to the one before it).
If a subsequence is non-decreasing, keep it. If not, discard it.
Finally, collect all the unique non-decreasing subsequences that you found.

Code Implementation

def find_non_decreasing_subsequences_brute_force(numbers):
    non_decreasing_subsequences = set()
    number_of_elements = len(numbers)

    for starting_index in range(number_of_elements):
        for i in range(starting_index, number_of_elements):
            current_subsequence = [numbers[starting_index]]
            find_subsequences(numbers, i + 1, current_subsequence, non_decreasing_subsequences)

    return list(non_decreasing_subsequences)

def find_subsequences(numbers, current_index, current_subsequence, non_decreasing_subsequences):
    # Base case: We've reached the end of the input list.
    if current_index == len(numbers):
        if len(current_subsequence) >= 2:
            non_decreasing_subsequences.add(tuple(current_subsequence))
        return

    # Recursive step: Include the current number in the subsequence
    # only if it's greater than or equal to the last element
    if numbers[current_index] >= current_subsequence[-1]:
        new_subsequence = current_subsequence + [numbers[current_index]]
        find_subsequences(numbers, current_index + 1, new_subsequence, non_decreasing_subsequences)

    # Ensure we check sequences without including the current number
    find_subsequences(numbers, current_index + 1, current_subsequence, non_decreasing_subsequences)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach generates all possible subsequences of the input array. An array of size n has 2^n possible subsequences, since each element can either be included or excluded from a subsequence. For each subsequence, we need to check if it is non-decreasing, which takes O(n) time in the worst case. However, generating all subsequences is the dominant factor, leading to a time complexity of O(2^n * n). Since 2^n grows much faster than n, we simplify the overall time complexity to O(2^n).

Space Complexity

O(2^N) – The algorithm explores all possible subsequences. In the worst case, for an input array of size N, there can be 2^N possible subsequences (each element can either be in or out of a subsequence). The algorithm needs to store all the unique non-decreasing subsequences. Therefore, the space required to store these subsequences in a data structure (e.g., a set to ensure uniqueness) can grow exponentially with the input size N. Hence, the space complexity is O(2^N).

Optimal Solution

Approach

The goal is to find all the increasing sequences within a given sequence. The optimal approach builds these increasing sequences step-by-step, avoiding redundant calculations by carefully managing which numbers are considered for each subsequence.

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

Start by considering each number in the input sequence as the potential beginning of an increasing sequence.
For each starting number, try to extend the sequence by adding numbers that are greater than or equal to the last number in the sequence.
Keep track of all the sequences you've built so far to avoid generating the same sequence multiple times. You can use a special tool to remember these sequences.
As you extend a sequence, make sure it has at least two numbers in it because single numbers by themself are not subsequences.
If you find a valid increasing sequence, add it to your collection of sequences.
Continue this process for each number in the input sequence to find all possible increasing sequences.

Code Implementation

def find_subsequences(nums):
    result = set()
    current_sequence = []

    def backtrack(index):
        # Ensure sequence has at least two elements
        if len(current_sequence) >= 2:
            result.add(tuple(current_sequence))

        if index == len(nums):
            return

        seen = set()
        for i in range(index, len(nums)):
            # Skip duplicates to avoid redundant subsequences.
            if nums[i] in seen:
                continue

            # Check if sequence is non-decreasing.
            if not current_sequence or nums[i] >= current_sequence[-1]:
                seen.add(nums[i])
                current_sequence.append(nums[i])
                backtrack(i + 1)
                current_sequence.pop()

    for i in range(len(nums)): # Begin sequence at each index
        backtrack(i)

    return [list(subsequence) for subsequence in result]

Big(O) Analysis

Time Complexity

O(n * 2^n) – The algorithm explores all possible subsequences of the input array of size n. In the worst-case scenario, where the input array is already sorted in non-decreasing order, each element can potentially be part of multiple non-decreasing subsequences. Extending the subsequences recursively could lead to examining a large number of combinations. Since a set of size n has 2^n subsets, and for each element, we are potentially checking almost all other elements to extend subsequences, the time complexity can approach O(n * 2^n) due to the exponential growth of subsequences being built and the need to avoid duplicates among them.

Space Complexity

O(2^N) – The algorithm uses a set (the special tool mentioned for remembering sequences) to store the unique non-decreasing subsequences found. In the worst-case scenario, where the input array is sorted in ascending order, each number can be part of multiple increasing subsequences. The maximum number of possible subsequences for an array of size N is 2^N, since each element can either be included or excluded from a subsequence. Thus, the space required to store all these subsequences can grow exponentially with the size of the input array. Note that each subsequence itself would take O(N) to store but the *number* of those sequences are the dominating factor.

Edge Cases

Case	How to Handle
Empty input array	Return an empty list immediately, as no subsequences can be formed.
Array with only one element	Return an empty list, as a non-decreasing subsequence requires at least two elements.
Array with all identical elements	The algorithm should generate all possible subsequences including duplicates, as they will be non-decreasing.
Array with a large number of duplicate elements	Use a set-based approach during recursion to avoid generating duplicate subsequences and improve performance.
Array with negative numbers	The algorithm should handle negative numbers correctly as part of the non-decreasing comparison.
Array with positive and negative numbers	The algorithm must correctly handle the comparisons between positive and negative numbers to form non-decreasing subsequences.
Maximum-sized input array causing stack overflow with recursion	Consider iterative approaches using bit manipulation or dynamic programming to avoid exceeding recursion depth limits.
Integer overflow during length calculations if applicable	Ensure length calculations do not cause integer overflow by using appropriate data types (e.g., long) where necessary.

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:

Can the input array contain duplicate numbers, and if so, are duplicates allowed within a non-decreasing subsequence?
What is the expected output if the input array is empty?
What is the data type of the numbers in the input array, and are there any constraints on the range of values?
Should the returned subsequences be sorted in any specific order?
Is the input array guaranteed to have at least one element?

Brute Force Solution

Approach

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

Start by considering a subsequence containing only the first element of the original sequence.
Then, consider all subsequences containing the first two elements, then the first three, and so on, until you've considered subsequences starting with the first element and using all other elements.
Next, repeat the process, but start with the second element as the beginning of the subsequence, and consider all combinations of elements after it.
Continue doing this, shifting the starting element one by one to the right, until you have considered starting with every possible element of the original sequence.
For each subsequence you consider, check if its elements are in non-decreasing order (each element is greater than or equal to the one before it).
If a subsequence is non-decreasing, keep it. If not, discard it.
Finally, collect all the unique non-decreasing subsequences that you found.

Code Implementation

def find_non_decreasing_subsequences_brute_force(numbers):
    non_decreasing_subsequences = set()
    number_of_elements = len(numbers)

    for starting_index in range(number_of_elements):
        for i in range(starting_index, number_of_elements):
            current_subsequence = [numbers[starting_index]]
            find_subsequences(numbers, i + 1, current_subsequence, non_decreasing_subsequences)

    return list(non_decreasing_subsequences)

def find_subsequences(numbers, current_index, current_subsequence, non_decreasing_subsequences):
    # Base case: We've reached the end of the input list.
    if current_index == len(numbers):
        if len(current_subsequence) >= 2:
            non_decreasing_subsequences.add(tuple(current_subsequence))
        return

    # Recursive step: Include the current number in the subsequence
    # only if it's greater than or equal to the last element
    if numbers[current_index] >= current_subsequence[-1]:
        new_subsequence = current_subsequence + [numbers[current_index]]
        find_subsequences(numbers, current_index + 1, new_subsequence, non_decreasing_subsequences)

    # Ensure we check sequences without including the current number
    find_subsequences(numbers, current_index + 1, current_subsequence, non_decreasing_subsequences)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by considering each number in the input sequence as the potential beginning of an increasing sequence.
For each starting number, try to extend the sequence by adding numbers that are greater than or equal to the last number in the sequence.
Keep track of all the sequences you've built so far to avoid generating the same sequence multiple times. You can use a special tool to remember these sequences.
As you extend a sequence, make sure it has at least two numbers in it because single numbers by themself are not subsequences.
If you find a valid increasing sequence, add it to your collection of sequences.
Continue this process for each number in the input sequence to find all possible increasing sequences.

Code Implementation

def find_subsequences(nums):
    result = set()
    current_sequence = []

    def backtrack(index):
        # Ensure sequence has at least two elements
        if len(current_sequence) >= 2:
            result.add(tuple(current_sequence))

        if index == len(nums):
            return

        seen = set()
        for i in range(index, len(nums)):
            # Skip duplicates to avoid redundant subsequences.
            if nums[i] in seen:
                continue

            # Check if sequence is non-decreasing.
            if not current_sequence or nums[i] >= current_sequence[-1]:
                seen.add(nums[i])
                current_sequence.append(nums[i])
                backtrack(i + 1)
                current_sequence.pop()

    for i in range(len(nums)): # Begin sequence at each index
        backtrack(i)

    return [list(subsequence) for subsequence in result]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return an empty list immediately, as no subsequences can be formed.
Array with only one element	Return an empty list, as a non-decreasing subsequence requires at least two elements.
Array with all identical elements	The algorithm should generate all possible subsequences including duplicates, as they will be non-decreasing.
Array with a large number of duplicate elements	Use a set-based approach during recursion to avoid generating duplicate subsequences and improve performance.
Array with negative numbers	The algorithm should handle negative numbers correctly as part of the non-decreasing comparison.
Array with positive and negative numbers	The algorithm must correctly handle the comparisons between positive and negative numbers to form non-decreasing subsequences.
Maximum-sized input array causing stack overflow with recursion	Consider iterative approaches using bit manipulation or dynamic programming to avoid exceeding recursion depth limits.
Integer overflow during length calculations if applicable	Ensure length calculations do not cause integer overflow by using appropriate data types (e.g., long) where necessary.