Count Special Subsequences

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ArraysDynamic Programming

You are given an array nums consisting of positive integers.

A special subsequence is defined as a subsequence of length 4, represented by indices (p, q, r, s), where p < q < r < s. This subsequence must satisfy the following conditions:

nums[p] * nums[r] == nums[q] * nums[s]
There must be at least one element between each pair of indices. In other words, q - p > 1, r - q > 1 and s - r > 1.

Return the number of different special subsequences in nums.

Example 1:

Input: nums = [1,2,3,4,3,6,1]

Output: 1

Explanation:

There is one special subsequence in nums.

(p, q, r, s) = (0, 2, 4, 6):
- This corresponds to elements (1, 3, 3, 1).
- nums[p] * nums[r] = nums[0] * nums[4] = 1 * 3 = 3
- nums[q] * nums[s] = nums[2] * nums[6] = 3 * 1 = 3

Example 2:

Input: nums = [3,4,3,4,3,4,3,4]

Output: 3

Explanation:

There are three special subsequences in nums.

(p, q, r, s) = (0, 2, 4, 6):
- This corresponds to elements (3, 3, 3, 3).
- nums[p] * nums[r] = nums[0] * nums[4] = 3 * 3 = 9
- nums[q] * nums[s] = nums[2] * nums[6] = 3 * 3 = 9
(p, q, r, s) = (1, 3, 5, 7):
- This corresponds to elements (4, 4, 4, 4).
- nums[p] * nums[r] = nums[1] * nums[5] = 4 * 4 = 16
- nums[q] * nums[s] = nums[3] * nums[7] = 4 * 4 = 16
(p, q, r, s) = (0, 2, 5, 7):
- This corresponds to elements (3, 3, 4, 4).
- nums[p] * nums[r] = nums[0] * nums[5] = 3 * 4 = 12
- nums[q] * nums[s] = nums[2] * nums[7] = 3 * 4 = 12

Constraints:

7 <= nums.length <= 1000
1 <= nums[i] <= 1000

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 is the range of values within the input array? Can they be negative or zero?
Can the input array be empty or null? What should I return in those cases?
What is considered a 'special subsequence' in this context? Are there specific properties it must satisfy that are not explicitly mentioned?
Are duplicates allowed within the input array, and if so, how do they affect the counting of special subsequences?
What should I return if no special subsequences exist in the input array?

Brute Force Solution

Approach

The brute force method is like trying out every possible arrangement until we find all the sequences that meet specific criteria. We explore all combinations, even the ones that don't make sense, until we find the special ones. It's exhaustive and guaranteed to work, but might take a long time.

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

Consider the first number in the given set of numbers.
Decide whether or not to include it in a possible sequence.
Then move to the next number in the set, again deciding whether or not to include it in the sequence.
Repeat this process for every single number in the set, creating all possible combinations of included and excluded numbers.
For each of these combinations, check if it follows the rules to be considered a special sequence. For example, does it start with 0, then have only 0s and 1s, and end with 2s?
If a combination follows the rules, count it as a special sequence.
After checking every single combination, add up the counts of all the special sequences you found.

Code Implementation

def count_special_subsequences_brute_force(numbers):
    number_of_numbers = len(numbers)
    special_subsequence_count = 0

    # Iterate through all possible subsequences using bit manipulation
    for i in range(2**number_of_numbers):
        subsequence = []

        for j in range(number_of_numbers):
            # Check if the j-th bit is set in i
            if (i >> j) & 1:
                subsequence.append(numbers[j])

        # Check if the subsequence is special
        if is_special_subsequence(subsequence):

            # Increment the count if it is
            special_subsequence_count += 1

    return special_subsequence_count

def is_special_subsequence(subsequence):
    if not subsequence:
        return False

    # Special subsequences must start with 0
    if subsequence[0] != 0:
        return False

    state = 0
    for number in subsequence:

        if state == 0:
            if number == 0:
                continue
            else:
                return False

        if state == 1:
            if number == 0 or number == 1:
                continue
            else:
                return False

        if state == 2:
            if number == 2:
                continue
            else:
                return False

        if number == 1 and state == 0:
            state = 1

        if number == 2 and (state == 0 or state == 1):

            # Transition to state 2 only when we encounter a 2
            state = 2

    if state == 2:
        return True

    return False

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves iterating through 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 check if it's a special subsequence. The cost of validating if a subsequence meets the criteria to be a special sequence is O(n) in the worst case, where n is the size of the input array. However, the dominating factor is generating and considering all 2^n subsequences. Therefore, the time complexity is O(2^n).

Space Complexity

O(N) – The brute force approach involves exploring all possible subsequences by making decisions (include or exclude) for each number in the input set. This decision-making process can be visualized as a binary tree of depth N, where N is the number of elements in the input. Although the plain english explanation doesn't explicitly store the decision tree, the implicit recursion stack necessary to explore all possible combinations can grow up to N levels deep, where each level corresponds to a number in the input. This results in an auxiliary space complexity proportional to the depth of the recursion, or O(N). The space complexity arises from keeping track of the call stack during the recursive exploration of all possible subsequences.

Optimal Solution

Approach

The key to solving this problem efficiently is to build valid subsequences step by step, keeping track of how many subsequences of each type you've created so far. We'll do this by processing each number in the original sequence only once and updating our counts accordingly.

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

Recognize that a 'special subsequence' can only contain 0s, 1s, and 2s, and must appear in that order.
Start with counts of zero for subsequences ending in '0', '0, 1', and '0, 1, 2'.
Go through the original sequence number by number.
If you encounter a '0', update the count of subsequences ending in '0' by adding 1 to it. This essentially means you've found one more '0' to start a new subsequence, and it also counts as a subsequence in and of itself.
If you encounter a '1', update the count of subsequences ending in '0, 1'. To do this, add the current count of subsequences ending in '0' to it. This means every existing subsequence ending in '0' can now be extended by appending a '1'.
If you encounter a '2', update the count of subsequences ending in '0, 1, 2'. To do this, add the current count of subsequences ending in '0, 1' to it. This means every existing subsequence ending in '0, 1' can now be extended by appending a '2'.
Since the numbers can get very large, you might need to take the result modulo some large number to prevent it from exceeding the memory limits.
Finally, after going through the entire sequence, the count of subsequences ending in '0, 1, 2' will be the answer: the total count of special subsequences.

Code Implementation

def count_special_subsequences(sequence):
    modulo = 10**9 + 7
    subsequence_ending_with_0 = 0
    subsequence_ending_with_0_1 = 0
    subsequence_ending_with_0_1_2 = 0

    for number in sequence:
        if number == 0:
            # Each new 0 can either start a subsequence or be appended.
            subsequence_ending_with_0 = (2 * subsequence_ending_with_0 + 1) % modulo

        elif number == 1:
            # Each new 1 extends existing '0' subsequences.
            subsequence_ending_with_0_1 = (2 * subsequence_ending_with_0_1 + subsequence_ending_with_0) % modulo

        elif number == 2:
            # Each new 2 extends existing '0, 1' subsequences.
            subsequence_ending_with_0_1_2 = (2 * subsequence_ending_with_0_1_2 + subsequence_ending_with_0_1) % modulo

    # The total number of '0, 1, 2' subsequences is the answer.
    return subsequence_ending_with_0_1_2

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input sequence once. For each element in the sequence (where the sequence size is denoted by n), it performs a fixed number of operations: checking if the element is 0, 1, or 2, and updating the corresponding count. The operations within the loop are constant time operations. Therefore, the time complexity is directly proportional to the size of the input sequence, resulting in a linear time complexity of O(n).

Space Complexity

O(1) – The algorithm uses three integer variables to store the counts of subsequences ending in '0', '0, 1', and '0, 1, 2' respectively. These variables consume a fixed amount of space regardless of the input sequence's length (N). Therefore, the auxiliary space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there are no subsequences.
Array contains only zeros	The number of special subsequences will depend on the allowed subsequences' rules; return 0 or calculate 2^(n-1) based on these rules if '0,1,2' is the subsequence.
Array with one element	Return 0 as a special subsequence of at least length 2 is not possible.
Array containing negative numbers (if subsequence numbers must be non-negative)	Filter out negative numbers before processing or return 0 if negative numbers are not allowed.
Integer overflow when calculating the count of subsequences	Use modulo arithmetic with a large prime number to prevent overflow.
Maximum sized input array (performance considerations)	Ensure the algorithm's time complexity is within acceptable bounds (e.g., O(n) or O(n log n)) to avoid timeouts.
No valid special subsequences exist in the input array	Return 0 when no special subsequence is found after processing the array.
Array contains extreme boundary values (Integer.MAX_VALUE, Integer.MIN_VALUE)	Ensure calculations involving these values do not cause overflow or unexpected behavior.

Null or empty input array

How to Handle:

Return 0, as there are no subsequences.

Array contains only zeros

How to Handle:

The number of special subsequences will depend on the allowed subsequences' rules; return 0 or calculate 2^(n-1) based on these rules if '0,1,2' is the subsequence.

Array with one element

How to Handle:

Return 0 as a special subsequence of at least length 2 is not possible.

Array containing negative numbers (if subsequence numbers must be non-negative)

How to Handle:

Filter out negative numbers before processing or return 0 if negative numbers are not allowed.

Integer overflow when calculating the count of subsequences

How to Handle:

Use modulo arithmetic with a large prime number to prevent overflow.

Maximum sized input array (performance considerations)

How to Handle:

Ensure the algorithm's time complexity is within acceptable bounds (e.g., O(n) or O(n log n)) to avoid timeouts.

No valid special subsequences exist in the input array

How to Handle:

Return 0 when no special subsequence is found after processing the array.

Array contains extreme boundary values (Integer.MAX_VALUE, Integer.MIN_VALUE)

How to Handle:

Ensure calculations involving these values do not cause overflow or unexpected behavior.