How do you solve Count Number of Special Subsequences on LeetCode?

Count Number of Special Subsequences (LeetCode #1955) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Subsequences must be in the order of 0s, then 1s, then 2s.

What is the brute force solution for Count Number of Special Subsequences?

The brute force approach for Count Number of Special Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the array and checking each one to see if it meets the criteria of being special. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Number of Special Subsequences?

Count Number of Special Subsequences uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Counting Subsequences.

What are the interview tips for Count Number of Special Subsequences?

Always clarify the problem requirements and constraints before diving into coding. Consider edge cases, such as arrays with only one type of number. Discuss your thought process and any potential optimizations with the interviewer.

What are common mistakes when solving Count Number of Special Subsequences?

Not considering the order of elements in subsequences. Overcounting subsequences by not properly managing counts.

#1955

Count Number of Special Subsequences

Hard

ArrayDynamic ProgrammingDynamic ProgrammingCounting Subsequences

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses dynamic programming to efficiently count the number of special subsequences without generating all possible subsequences. We keep track of the counts of subsequences ending with 0s, 1s, and 2s, allowing us to build upon previous counts.

⚙️

Algorithm

4 steps

1Step 1: Initialize counts for subsequences ending with 0s, 1s, and 2s.
2Step 2: Iterate through the array, updating counts based on the current number (0, 1, or 2).
3Step 3: For each 0, increment the count of subsequences ending with 0s. For each 1, add the count of subsequences ending with 0s to the count of subsequences ending with 1s. For each 2, add the count of subsequences ending with 1s to the count of subsequences ending with 2s.
4Step 4: The final count of special subsequences is the count of subsequences ending with 2s.

solution.py12 lines

1# Full working Python code
2def count_special_subsequences(nums):
3    MOD = 10**9 + 7
4    count_0 = count_1 = count_2 = 0
5    for num in nums:
6        if num == 0:
7            count_0 = (count_0 * 2 + 1) % MOD
8        elif num == 1:
9            count_1 = (count_1 * 2 + count_0) % MOD
10        elif num == 2:
11            count_2 = (count_2 * 2 + count_1) % MOD
12    return count_2

ℹ

Complexity note: This complexity is linear because we only traverse the array once, updating counts based on the current number without needing extra space for subsequences.

1Subsequences must be in the order of 0s, then 1s, then 2s.
2Dynamic programming can help efficiently count valid subsequences without generating them.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.