Count Number of Special Subsequences

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ArraysDynamic Programming

A sequence is special if it consists of a positive number of 0s, followed by a positive number of 1s, then a positive number of 2s.

For example, [0,1,2] and [0,0,1,1,1,2] are special.
In contrast, [2,1,0], [1], and [0,1,2,0] are not special.

Given an array nums (consisting of only integers 0, 1, and 2), return the number of different subsequences that are special. Since the answer may be very large, return it modulo 10⁹ + 7.

A subsequence of an array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements. Two subsequences are different if the set of indices chosen are different.

Example 1:

Input: nums = [0,1,2,2]
Output: 3
Explanation: The special subsequences are bolded [0,1,2,2], [0,1,2,2], and [0,1,2,2].

Example 2:

Input: nums = [2,2,0,0]
Output: 0
Explanation: There are no special subsequences in [2,2,0,0].

Example 3:

Input: nums = [0,1,2,0,1,2]
Output: 7
Explanation: The special subsequences are bolded:
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]
- [0,1,2,0,1,2]

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 2

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 length of the input array?
Can the input array contain negative numbers, zeros, or non-integer values?
What constitutes a "special subsequence"? Can you provide more specific criteria beyond the general definition?
If no special subsequence exists, what should I return?
Are there any memory constraints I should be aware of in addition to time complexity considerations?

Brute Force Solution

Approach

To count special subsequences, the brute force method explores every possible subsequence. We generate each subsequence and then check if it meets the 'special' criteria. Finally, we count the subsequences that qualify.

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

Start by considering every possible combination of elements from the original sequence.
For each of these combinations, check if it follows the rules of a 'special' subsequence. This means verifying if the sequence has the specific pattern or characteristics we're looking for.
If a combination satisfies all the rules, mark it as a valid 'special' subsequence.
After checking all the combinations, count how many were marked as valid. This count will be the answer to the problem.

Code Implementation

def count_number_special_subsequences_brute_force(sequence):
    number_of_elements = len(sequence)
    count = 0

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

        # Check if the subsequence is special.
        if is_special(subsequence):
            count += 1

    return count

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

    # A special sequence must start with 0s,
    # followed by 1s, and end with 2s
    state = 0  # 0: expecting 0, 1: expecting 1, 2: expecting 2

    for number in subsequence:
        if state == 0:
            if number == 0:
                continue
            elif number == 1:
                state = 1
            else:
                return False
        elif state == 1:
            if number == 1:
                continue
            elif number == 2:
                state = 2
            else:
                return False
        else:
            if number == 2:
                continue
            else:
                return False

    # The sequence must end in state 2.
    return state == 2

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach generates every possible subsequence. For an input sequence of size n, there are 2^n possible subsequences. For each subsequence, we need to check if it's 'special', which involves iterating through the subsequence to verify its properties. In the worst case, this verification takes O(n) time where n is the length of the original sequence because the subsequence length can be at most n. Therefore, the total time complexity is O(2^n * n).

Space Complexity

O(1) – The brute force method, as described, explores every possible subsequence. While generating each subsequence, it does not explicitly mention creating new data structures to hold these subsequences. Checking the subsequence's validity does not imply additional data structures either. The 'marking' and 'counting' operations can be done using a single counter variable. Therefore, the auxiliary space used remains constant regardless of the input sequence's length N.

Optimal Solution

Approach

The problem asks us to count special subsequences. Instead of checking every single subsequence, we can build our count dynamically by tracking how many valid subsequences end with each possible number (0, 1, or 2). This allows us to calculate the count efficiently based on previous results.

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

Imagine we are building the special subsequences one number at a time.
We'll keep track of three counts: how many subsequences end in 0, how many end in 1, and how many end in 2.
Go through the input sequence, number by number.
If the current number is 0, we can either start a new subsequence with it, or add it to existing subsequences that end in 0. This updates our count of subsequences ending in 0.
If the current number is 1, we can add it to any existing subsequence that ends in 0, or simply start a new subsequence consisting of just '1'. This updates the count of subsequences ending in 1.
If the current number is 2, we can add it to any existing subsequence that ends in 1. This updates the count of subsequences ending in 2.
After processing all the numbers, the count of subsequences ending in 2 will be our final answer because any special subsequence must end with 2.

Code Implementation

def count_special_subsequences(sequence):
    end_in_zero_count = 0
    end_in_one_count = 0
    end_in_two_count = 0

    for number in sequence:

        if number == 0:
            # Either start a new one or add to existing
            end_in_zero_count = (2 * end_in_zero_count + 1) % 1000000007

        elif number == 1:
            # Special subsequence must start with 0, add this one after 0
            end_in_one_count = (2 * end_in_one_count + end_in_zero_count) % 1000000007

        else:
            # A special subsequence must end with 2, add this one after 1
            end_in_two_count = (2 * end_in_two_count + end_in_one_count) % 1000000007

    return end_in_two_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input sequence of numbers (let's say its length is n) exactly once. Inside the loop, we perform a constant number of operations (updating the counts of subsequences ending in 0, 1, or 2 based on the current number). Therefore, the time complexity is directly proportional to the number of elements in the input sequence, resulting in O(n).

Space Complexity

O(1) – The algorithm uses three variables to store the counts of subsequences ending in 0, 1, and 2 respectively. The number of these variables does not depend on the input size, N, which represents the length of the input sequence. Therefore, the auxiliary space required is constant, independent of the input size. The space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there are no subsequences in an empty array.
Array with only one element	Return 0, a single element cannot form a special subsequence as defined by the problem (0, 1, 2).
Array with all zeros	Calculate 2^n - 1 representing the number of subsequences filled with zeros, but using modulo to avoid overflow.
Array with no '0', '1', or '2' elements	Return 0, as no special subsequence can be formed.
Large array size leading to potential integer overflow during count calculation	Use modulo operator during calculations to prevent integer overflow.
Array with a mix of 0s, 1s and 2s in a specific order that maximizes count	The dynamic programming approach inherently handles this case by considering all possible transitions.
Maximum sized input array	The dynamic programming solution scales linearly with input array size which should be sufficient.
Array with extremely large numbers of 0s or 1s leading to overflow of intermediate calculations even with modulo	Ensure the modulo value is sufficiently large to avoid overflow in the intermediate products during count calculation.

Null or empty input array

How to Handle:

Return 0, as there are no subsequences in an empty array.

Array with only one element

How to Handle:

Return 0, a single element cannot form a special subsequence as defined by the problem (0, 1, 2).

Array with all zeros

How to Handle:

Calculate 2^n - 1 representing the number of subsequences filled with zeros, but using modulo to avoid overflow.

Array with no '0', '1', or '2' elements

How to Handle:

Return 0, as no special subsequence can be formed.

Large array size leading to potential integer overflow during count calculation

How to Handle:

Use modulo operator during calculations to prevent integer overflow.

Array with a mix of 0s, 1s and 2s in a specific order that maximizes count

How to Handle:

The dynamic programming approach inherently handles this case by considering all possible transitions.

Maximum sized input array

How to Handle:

The dynamic programming solution scales linearly with input array size which should be sufficient.

Array with extremely large numbers of 0s or 1s leading to overflow of intermediate calculations even with modulo

How to Handle:

Ensure the modulo value is sufficiently large to avoid overflow in the intermediate products during count calculation.