LeetCode #1955 — HARD

Count Number of Special Subsequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[0,1,2,2]

Example 2

[2,2,0,0]

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Can we first solve a simpler problem? Counting the number of subsequences with 1s followed by 0s.
How can we keep track of the partially matched subsequences to help us find the answer?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1955: Count Number of Special Subsequences
class Solution {
    public int countSpecialSubsequences(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int[][] f = new int[n][3];
        f[0][0] = nums[0] == 0 ? 1 : 0;
        for (int i = 1; i < n; ++i) {
            if (nums[i] == 0) {
                f[i][0] = (2 * f[i - 1][0] % mod + 1) % mod;
                f[i][1] = f[i - 1][1];
                f[i][2] = f[i - 1][2];
            } else if (nums[i] == 1) {
                f[i][0] = f[i - 1][0];
                f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1] % mod) % mod;
                f[i][2] = f[i - 1][2];
            } else {
                f[i][0] = f[i - 1][0];
                f[i][1] = f[i - 1][1];
                f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2] % mod) % mod;
            }
        }
        return f[n - 1][2];
    }
}

// Accepted solution for LeetCode #1955: Count Number of Special Subsequences
func countSpecialSubsequences(nums []int) int {
	const mod = 1e9 + 7
	n := len(nums)
	f := make([][3]int, n)
	if nums[0] == 0 {
		f[0][0] = 1
	}
	for i := 1; i < n; i++ {
		if nums[i] == 0 {
			f[i][0] = (2*f[i-1][0] + 1) % mod
			f[i][1] = f[i-1][1]
			f[i][2] = f[i-1][2]
		} else if nums[i] == 1 {
			f[i][0] = f[i-1][0]
			f[i][1] = (f[i-1][0] + 2*f[i-1][1]) % mod
			f[i][2] = f[i-1][2]
		} else {
			f[i][0] = f[i-1][0]
			f[i][1] = f[i-1][1]
			f[i][2] = (f[i-1][1] + 2*f[i-1][2]) % mod
		}
	}
	return f[n-1][2]
}

# Accepted solution for LeetCode #1955: Count Number of Special Subsequences
class Solution:
    def countSpecialSubsequences(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        n = len(nums)
        f = [[0] * 3 for _ in range(n)]
        f[0][0] = nums[0] == 0
        for i in range(1, n):
            if nums[i] == 0:
                f[i][0] = (2 * f[i - 1][0] + 1) % mod
                f[i][1] = f[i - 1][1]
                f[i][2] = f[i - 1][2]
            elif nums[i] == 1:
                f[i][0] = f[i - 1][0]
                f[i][1] = (f[i - 1][0] + 2 * f[i - 1][1]) % mod
                f[i][2] = f[i - 1][2]
            else:
                f[i][0] = f[i - 1][0]
                f[i][1] = f[i - 1][1]
                f[i][2] = (f[i - 1][1] + 2 * f[i - 1][2]) % mod
        return f[n - 1][2]

// Accepted solution for LeetCode #1955: Count Number of Special Subsequences
/**
 * [1955] Count Number of Special Subsequences
 *
 * 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 10^9 + 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 [<u>0</u>,<u>1</u>,<u>2</u>,2], [<u>0</u>,<u>1</u>,2,<u>2</u>], and [<u>0</u>,<u>1</u>,<u>2</u>,<u>2</u>].
 *
 * 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:
 * - [<u>0</u>,<u>1</u>,<u>2</u>,0,1,2]
 * - [<u>0</u>,<u>1</u>,2,0,1,<u>2</u>]
 * - [<u>0</u>,<u>1</u>,<u>2</u>,0,1,<u>2</u>]
 * - [<u>0</u>,<u>1</u>,2,0,<u>1</u>,<u>2</u>]
 * - [<u>0</u>,1,2,<u>0</u>,<u>1</u>,<u>2</u>]
 * - [<u>0</u>,1,2,0,<u>1</u>,<u>2</u>]
 * - [0,1,2,<u>0</u>,<u>1</u>,<u>2</u>]
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 10^5
 * 	0 <= nums[i] <= 2
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/count-number-of-special-subsequences/
// discuss: https://leetcode.com/problems/count-number-of-special-subsequences/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_special_subsequences(nums: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1955_example_1() {
        let nums = vec![0, 1, 2, 2];

        let result = 3;

        assert_eq!(Solution::count_special_subsequences(nums), result);
    }

    #[test]
    #[ignore]
    fn test_1955_example_2() {
        let nums = vec![2, 2, 0, 0];

        let result = 0;

        assert_eq!(Solution::count_special_subsequences(nums), result);
    }

    #[test]
    #[ignore]
    fn test_1955_example_3() {
        let nums = vec![0, 1, 2, 0, 1, 2];

        let result = 7;

        assert_eq!(Solution::count_special_subsequences(nums), result);
    }
}

// Accepted solution for LeetCode #1955: Count Number of Special Subsequences
function countSpecialSubsequences(nums: number[]): number {
    const mod = 1e9 + 7;
    const n = nums.length;
    const f: number[][] = Array(n)
        .fill(0)
        .map(() => Array(3).fill(0));
    f[0][0] = nums[0] === 0 ? 1 : 0;
    for (let i = 1; i < n; ++i) {
        if (nums[i] === 0) {
            f[i][0] = (((2 * f[i - 1][0]) % mod) + 1) % mod;
            f[i][1] = f[i - 1][1];
            f[i][2] = f[i - 1][2];
        } else if (nums[i] === 1) {
            f[i][0] = f[i - 1][0];
            f[i][1] = (f[i - 1][0] + ((2 * f[i - 1][1]) % mod)) % mod;
            f[i][2] = f[i - 1][2];
        } else {
            f[i][0] = f[i - 1][0];
            f[i][1] = f[i - 1][1];
            f[i][2] = (f[i - 1][1] + ((2 * f[i - 1][2]) % mod)) % mod;
        }
    }
    return f[n - 1][2];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.