Hard: LeetCode 2338. Count the Number of Ideal Arrays. Python3. Sieve & Combinatorics & Prime‑power.

🚀 Count the Number of Ideal Arrays – Sieve & Combinatorics Explained!
Just tackled Count the Number of Ideal Arrays on LeetCode! This problem asks us to count all length‑n arrays where each element divides the next, and I solved it using a smart blend of Sieve of Eratosthenes for fast prime factorization and stars‑and‑bars combinatorics for counting distributions.

🔹 Key Strategy:

✅ Precompute each number’s smallest prime divisor with a sieve for O(1) factor queries

✅ Use “stars and bars” (C(n + k − 1, k)) to count ways to distribute k prime powers across n slots

✅ Multiply independent prime choices per value and sum mod 1e9+7


Story
Imagine you’re building a tower of colorful blocks, one on top of the other. You have n layers to stack, and at each layer you can choose any block numbered from 1 to maxValue. But there’s a rule: the number on each block must perfectly divide the number on the block above it, so the tower doesn’t wobble. How many stable towers can you build?

First, you learn the prime makeup of every block size using a “block sieve,” so you know which prime pieces compose each block. Next, for each prime piece you decide how many times it can appear in the tower by turning that into a simple combinatorial count—it’s like choosing where to place identical Lego bricks among your layers. Finally, you multiply those prime‑brick choices together for each starting block and add them up. By mixing a classic sieve, stars‑and‑bars math, and prime factor jumps, you discover the total number of wobble‑free towers in just one pass!

🧑‍💻 Dive into the full breakdown here:

📌 LeetCode Solution: https://leetcode.com/problems/count-the-nu...

📌 GitHub Repository: https://github.com/RuslanTsykaliak/LeetCod...

📌 LinkedIn Post: https://www.linkedin.com/posts/ruslan-tsyk...

Understanding combinatorics and sieve tricks is crucial for tackling high‑performance counting problems. Would you reach for these techniques next time? Yes or No? 🤔

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