Introduction

In this blog post, we will explore how to solve the LeetCode problem "1955" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Dynamic Programming (DP): Maintain counts of subsequences ending with 0, 0 followed by 1, and 0 followed by 1 followed by 2.
• Iterative Update: Iterate through the input array, updating the counts based on the current element. If we see a 0, we can extend previous 0-ending subsequences or start a new subsequence. If we see a 1, we extend 0-ending subsequences. If we see a 2, we extend 01-ending subsequences.

• Modulo Arithmetic: Apply the modulo operator to prevent overflow, since the number of subsequences can be very large.

• Time Complexity: O(n), where n is the length of the input array. Space Complexity: O(1).

Code

    def count_special_subsequences(nums):
    """
    Counts the number of special subsequences in the given array.

    Args:
        nums: A list of integers consisting of 0, 1, and 2.

    Returns:
        The number of different subsequences that are special, modulo 10^9 + 7.
    """

    MOD = 10**9 + 7
    count0 = 0
    count01 = 0
    count012 = 0

    for num in nums:
        if num == 0:
            count0 = (2 * count0 + 1) % MOD  # Extend existing or start new
        elif num == 1:
            count01 = (2 * count01 + count0) % MOD  # Extend existing 01 with this 1, or extend 0 sequences with 1
        else:  # num == 2
            count012 = (2 * count012 + count01) % MOD  # Extend existing 012 with this 2, or extend 01 sequences with 2

    return count012