Is Count Number of Special Subsequences easy or hard?

Count Number of Special Subsequences is rated Hard because it requires recognizing a subsequence pattern and modeling it with dynamic programming state transitions. Once the three-state DP pattern is understood, the implementation becomes straightforward.

Count Number of Special Subsequences Python/Java solution

Both Python and Java implementations follow the same DP idea. Maintain three variables for counts of 0, 01, and 012 subsequences and update them during a single pass through the array. The logic is identical across languages with only syntax differences.

How to solve Count Number of Special Subsequences in O(n)?

Maintain three counters: count0, count01, and count012. For each element, update the counters based on whether the value is 0, 1, or 2. A 0 doubles existing 0-only subsequences, a 1 extends 0 subsequences into 01 subsequences, and a 2 extends 01 subsequences into complete 012 subsequences.

What is the best approach for Count Number of Special Subsequences?

The optimal approach uses dynamic programming with three state counters: sequences of 0s, sequences of 0s followed by 1s, and sequences of 0s→1s→2s. As you iterate through the array, each number updates one of these states. This solution runs in O(n) time and O(1) space and is the standard interview approach.

Is Count Number of Special Subsequences asked at Google/Amazon/Meta?

Variants of subsequence counting and dynamic programming state transition problems appear frequently in interviews at companies like Google, Amazon, and Meta. Problems involving pattern-based subsequences and combinational counting are common in senior-level coding rounds.

What data structure is used in Count Number of Special Subsequences?

The solution primarily uses dynamic programming with simple integer counters rather than complex data structures. The input is processed as an array while maintaining three state variables that represent different subsequence stages.

What is the time complexity of Count Number of Special Subsequences?

The optimal solution runs in O(n) time because the array is scanned once and each element performs constant-time state updates. Space complexity is O(1) since only three counters are stored instead of a full DP table.

Description Hints 2 Solution Video 14:12

1955. Count Number of Special Subsequences

Hard53.4% acceptance1 company

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

Asked at 1 company

Amazon

Topics

Array Dynamic Programming

Hints open one at a time

Hint 1

Can we first solve a simpler problem? Counting the number of subsequences with 1s followed by 0s.

Hint 2

How can we keep track of the partially matched subsequences to help us find the answer?

14:12

Still stuck?Read 6 approaches, or watch a walkthrough.

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1class Solution:
2    def countSpecialSubsequences(self, nums: List[int]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[0,1,2,2]

Input

[2,2,0,0]

Input

[0,1,2,0,1,2]

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