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Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

  1. Dynamic Programming Approach - Complex approach
  2. Dynamic Programming with Modular Arithmetic - Complex approach
  3. Dynamic Programming with Array - Complex approach

Approach 1: Dynamic Programming Approach

The key insight for solving this problem is to use dynamic programming to keep track of the number of subsequences ending with 0, 1, and 2 at each position.

Let's define three variables:

  • count0: Number of subsequences ending with 0
  • count1: Number of subsequences ending with 1 that have a 0 before the 1
  • count2: Number of subsequences ending with 2 that have a 0 and a 1 before the 2 (in that order)

As we iterate through the array, we update these counts based on the current element.

Algorithm:

  1. Initialize count0 = count1 = count2 = 0.
  2. Iterate through the array from left to right:
  3. a. If the current element is 0:
  4. i. Update count0 = 2 * count0 + 1 (double the existing count0 and add 1 for the new subsequence consisting of just this 0).
  5. b. If the current element is 1:
  6. i. Update count1 = 2 * count1 + count0 (double the existing count1 and add count0 for the new subsequences formed by appending this 1 to existing subsequences ending with 0).
  7. c. If the current element is 2:
  8. i. Update count2 = 2 * count2 + count1 (double the existing count2 and add count1 for the new subsequences formed by appending this 2 to existing subsequences ending with 1).
  9. Return count2 as the final answer.

Time Complexity:

O(n)

Where n is the length of the array. We iterate through the array once, and each operation takes constant time.

Space Complexity:

O(1)

We only use three variables (count0, count1, count2) regardless of the input size.

Approach 2: Dynamic Programming with Modular Arithmetic

Since the answer may be very large, we need to use modular arithmetic to avoid overflow. We'll use the same approach as before, but apply the modulo operation at each step.

This approach is essentially the same as the previous one, but with modular arithmetic applied to handle large numbers.

Algorithm:

  1. Initialize count0 = count1 = count2 = 0 and MOD = 10^9 + 7.
  2. Iterate through the array from left to right:
  3. a. If the current element is 0:
  4. i. Update count0 = (2 * count0 + 1) % MOD.
  5. b. If the current element is 1:
  6. i. Update count1 = (2 * count1 + count0) % MOD.
  7. c. If the current element is 2:
  8. i. Update count2 = (2 * count2 + count1) % MOD.
  9. Return count2 as the final answer.

Time Complexity:

O(n)

Where n is the length of the array. We iterate through the array once, and each operation takes constant time.

Space Complexity:

O(1)

We only use three variables (count0, count1, count2) regardless of the input size.

Approach 3: Dynamic Programming with Array

An alternative approach is to use an array to keep track of the counts for each digit. This makes the code more general and easier to extend if we need to handle more than three digits.

Let's define an array dp where dp[i] represents the number of subsequences ending with digit i.

Algorithm:

  1. Initialize an array dp of size 3, with all elements set to 0, and MOD = 10^9 + 7.
  2. Iterate through the array from left to right:
  3. a. For each element nums[i]:
  4. i. If nums[i] = 0, update dp[0] = (2 * dp[0] + 1) % MOD.
  5. ii. If nums[i] = 1, update dp[1] = (2 * dp[1] + dp[0]) % MOD.
  6. iii. If nums[i] = 2, update dp[2] = (2 * dp[2] + dp[1]) % MOD.
  7. Return dp[2] as the final answer.

Time Complexity:

O(n)

Where n is the length of the array. We iterate through the array once, and each operation takes constant time.

Space Complexity:

O(1)

We use an array of fixed size (3) regardless of the input size.

Pseudocode

solution.pseudo
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function countSpecialSubsequences(nums):
    count0 = 0
    count1 = 0
    count2 = 0
 
    for i from 0 to length of nums - 1:
        if nums[i] == 0:
            count0 = 2 * count0 + 1
        else if nums[i] == 1:
            count1 = 2 * count1 + count0
        else if nums[i] == 2:
            count2 = 2 * count2 + count1
 
    return count2
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