How do you solve Number of Stable Subsequences on LeetCode?

Number of Stable Subsequences (LeetCode #3686) 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 of length 1 and 2 are always stable.

What is the time complexity of Number of Stable Subsequences?

The optimal solution for Number of Stable Subsequences runs in O(n) time and uses O(1) space. We only traverse the array once, maintaining counts without extra space for subsequences.

What is the brute force solution for Number of Stable Subsequences?

The brute force approach for Number of Stable Subsequences is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subsequences and check each for stability. This method is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

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

Number of Stable Subsequences uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Combinatorics.

What are the interview tips for Number of Stable Subsequences?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Number of Stable Subsequences?

Not considering subsequences of length 1 and 2 as valid. Incorrectly counting subsequences due to parity checks.

#3686

Number of Stable Subsequences

Hard

ArrayDynamic ProgrammingDynamic ProgrammingCombinatorics

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)

Use dynamic programming to count valid subsequences while maintaining the parity constraints. This avoids generating all subsequences explicitly.

⚙️

Algorithm

3 steps

1Step 1: Initialize dp arrays to count subsequences ending with odd and even numbers.
2Step 2: Iterate through the nums array, updating counts based on the last two elements' parity.
3Step 3: Return the total count of stable subsequences.

solution.py11 lines

1def stable_subsequences(nums):
2    mod = 10**9 + 7
3    dp_odd, dp_even = 0, 0
4    total = 0
5    for num in nums:
6        if num % 2 == 0:
7            dp_even = (dp_even + 1 + dp_odd) % mod
8        else:
9            dp_odd = (dp_odd + 1 + dp_even) % mod
10        total = (total + dp_odd + dp_even) % mod
11    return total

ℹ

Complexity note: We only traverse the array once, maintaining counts without extra space for subsequences.

1Subsequences of length 1 and 2 are always stable.
2Only adding a third element of the same parity invalidates the subsequence.

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