LeetCode #3686 — HARD

Number of Stable Subsequences

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

The Problem

Problem Statement

You are given an integer array nums.

A subsequence is stable if it does not contain three consecutive elements with the same parity when the subsequence is read in order (i.e., consecutive inside the subsequence).

Return the number of stable subsequences.

Since the answer may be too large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,3,5]

Output: 6

Explanation:

Stable subsequences are [1], [3], [5], [1, 3], [1, 5], and [3, 5].
Subsequence [1, 3, 5] is not stable because it contains three consecutive odd numbers. Thus, the answer is 6.

Example 2:

Input: nums = [2,3,4,2]

Output: 14

Explanation:

The only subsequence that is not stable is [2, 4, 2], which contains three consecutive even numbers.
All other subsequences are stable. Thus, the answer is 14.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A subsequence is stable if it does not contain three consecutive elements with the same parity when the subsequence is read in order (i.e., consecutive inside the subsequence). Return the number of stable subsequences. Since the answer may be too large, return it modulo 109 + 7.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

[1,3,5]

Example 2

[2,3,4,2]

Step 02

Core Insight

What unlocks the optimal approach

Any subsequence of length 1 or 2 is always stable.
A subsequence becomes invalid only if you add a third consecutive element of the same parity.
Use DP tracking the last element’s parity and how many consecutive of that parity you have (1 or 2).
For each new number, either start a new subsequence, extend with same parity (if <code>count < 2</code>), or extend with different parity (reset count = 1).

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 #3686: Number of Stable Subsequences
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3686: Number of Stable Subsequences
// package main
// 
// // https://space.bilibili.com/206214
// func countStableSubsequences(nums []int) int {
// 	const mod = 1_000_000_007
// 	f := [2][2]int{}
// 	for _, x := range nums {
// 		x %= 2
// 		f[x][1] = (f[x][1] + f[x][0]) % mod
// 		f[x][0] = (f[x][0] + f[x^1][0] + f[x^1][1] + 1) % mod
// 	}
// 	return (f[0][0] + f[0][1] + f[1][0] + f[1][1]) % mod
// }

// Accepted solution for LeetCode #3686: Number of Stable Subsequences
package main

// https://space.bilibili.com/206214
func countStableSubsequences(nums []int) int {
	const mod = 1_000_000_007
	f := [2][2]int{}
	for _, x := range nums {
		x %= 2
		f[x][1] = (f[x][1] + f[x][0]) % mod
		f[x][0] = (f[x][0] + f[x^1][0] + f[x^1][1] + 1) % mod
	}
	return (f[0][0] + f[0][1] + f[1][0] + f[1][1]) % mod
}

# Accepted solution for LeetCode #3686: Number of Stable Subsequences
#
# @lc app=leetcode id=3686 lang=python3
#
# [3686] Number of Stable Subsequences
#

# @lc code=start
class Solution:
    def countStableSubsequences(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        
        # dp[0]: count of stable subsequences ending with 1 even number
        # dp[1]: count of stable subsequences ending with 2 even numbers
        # dp[2]: count of stable subsequences ending with 1 odd number
        # dp[3]: count of stable subsequences ending with 2 odd numbers
        dp = [0] * 4
        
        for x in nums:
            if x % 2 == 0:
                # Current number is Even
                # Can form new subsequences ending in 2 evens by appending to those ending in 1 even
                added_to_1_even = dp[0]
                
                # Can form new subsequences ending in 1 even by appending to those ending in odds
                # Plus the single element subsequence [x]
                added_to_odds = (dp[2] + dp[3] + 1) % MOD
                
                dp[1] = (dp[1] + added_to_1_even) % MOD
                dp[0] = (dp[0] + added_to_odds) % MOD
            else:
                # Current number is Odd
                # Can form new subsequences ending in 2 odds by appending to those ending in 1 odd
                added_to_1_odd = dp[2]
                
                # Can form new subsequences ending in 1 odd by appending to those ending in evens
                # Plus the single element subsequence [x]
                added_to_evens = (dp[0] + dp[1] + 1) % MOD
                
                dp[3] = (dp[3] + added_to_1_odd) % MOD
                dp[2] = (dp[2] + added_to_evens) % MOD
                
        return sum(dp) % MOD
# @lc code=end

// Accepted solution for LeetCode #3686: Number of Stable Subsequences
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3686: Number of Stable Subsequences
// package main
// 
// // https://space.bilibili.com/206214
// func countStableSubsequences(nums []int) int {
// 	const mod = 1_000_000_007
// 	f := [2][2]int{}
// 	for _, x := range nums {
// 		x %= 2
// 		f[x][1] = (f[x][1] + f[x][0]) % mod
// 		f[x][0] = (f[x][0] + f[x^1][0] + f[x^1][1] + 1) % mod
// 	}
// 	return (f[0][0] + f[0][1] + f[1][0] + f[1][1]) % mod
// }

// Accepted solution for LeetCode #3686: Number of Stable Subsequences
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3686: Number of Stable Subsequences
// package main
// 
// // https://space.bilibili.com/206214
// func countStableSubsequences(nums []int) int {
// 	const mod = 1_000_000_007
// 	f := [2][2]int{}
// 	for _, x := range nums {
// 		x %= 2
// 		f[x][1] = (f[x][1] + f[x][0]) % mod
// 		f[x][0] = (f[x][0] + f[x^1][0] + f[x^1][1] + 1) % mod
// 	}
// 	return (f[0][0] + f[0][1] + f[1][0] + f[1][1]) % mod
// }

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 × m)

Space

O(n × m)

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.