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‎README.md

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| # | Title | Difficulty | Tag | Time, ms | Time, %
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|------|----------------|-------------|-------------|----------|--------
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| 3626 |[Find Stores with Inventory Imbalance](src/main/kotlin/g3601_3700/s3626_find_stores_with_inventory_imbalance)| Medium || 516 | 100.00
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| 3625 |[Count Number of Trapezoids II](src/main/kotlin/g3601_3700/s3625_count_number_of_trapezoids_ii)| Hard | Array, Hash_Table, Math, Geometry, Weekly_Contest_459 | 377 | 100.00
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| 3624 |[Number of Integers With Popcount-Depth Equal to K II](src/main/kotlin/g3601_3700/s3624_number_of_integers_with_popcount_depth_equal_to_k_ii)| Hard | Array, Segment_Tree, Weekly_Contest_459 | 38 | 100.00
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| 3623 |[Count Number of Trapezoids I](src/main/kotlin/g3601_3700/s3623_count_number_of_trapezoids_i)| Medium | Array, Hash_Table, Math, Geometry, Weekly_Contest_459 | 58 | 68.00

‎src/main/kotlin/g3601_3700/s3620_network_recovery_pathways/readme.md

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q.offer(i)
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}
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}
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while (!q.isEmpty()) {
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while (q.isNotEmpty()) {
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val u = q.poll()
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ts.add(u)
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for (v in g[u]) {

‎src/main/kotlin/g3601_3700/s3626_find_stores_with_inventory_imbalance/readme.md

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[![](https://img.shields.io/github/stars/javadev/LeetCode-in-Kotlin?label=Stars&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin)
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[![](https://img.shields.io/github/forks/javadev/LeetCode-in-Kotlin?label=Fork%20me%20on%20GitHub%20&style=flat-square)](https://github.com/javadev/LeetCode-in-Kotlin/fork)
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## 3626\. Find Stores with Inventory Imbalance
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Medium
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Table: `stores`
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+------------+----------+
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| Column Name| Type |
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+------------+----------+
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| store_id | int |
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| store_name | varchar |
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| location | varchar |
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+------------+----------+
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store_id is the unique identifier for this table.
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Each row contains information about a store and its location.
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Table: `inventory`
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+--------------+----------+
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| Column Name | Type |
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+--------------+----------+
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| inventory_id | int |
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| store_id | int |
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| product_name | varchar |
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| quantity | int |
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| price | decimal |
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+--------------+----------+
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inventory_id is the unique identifier for this table.
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Each row represents the inventory of a specific product at a specific store.
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Write a solution to find stores that have **inventory imbalance** - stores where the most expensive product has lower stock than the cheapest product.
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* For each store, identify the **most expensive product** (highest price) and its quantity
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* For each store, identify the **cheapest product** (lowest price) and its quantity
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* A store has inventory imbalance if the most expensive product's quantity is **less than** the cheapest product's quantity
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* Calculate the **imbalance ratio** as (cheapest\_quantity / most\_expensive\_quantity)
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* **Round** the imbalance ratio to **2** decimal places
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* Only include stores that have **at least** `3` **different products**
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Return _the result table ordered by imbalance ratio in **descending** order, then by store name in **ascending** order_.
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The result format is in the following example.
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**Example:**
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**Input:**
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stores table:
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+----------+----------------+-------------+
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| store_id | store_name | location |
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+----------+----------------+-------------+
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| 1 | Downtown Tech | New York |
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| 2 | Suburb Mall | Chicago |
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| 3 | City Center | Los Angeles |
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| 4 | Corner Shop | Miami |
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| 5 | Plaza Store | Seattle |
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+----------+----------------+-------------+
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inventory table:
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+--------------+----------+--------------+----------+--------+
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| inventory_id | store_id | product_name | quantity | price |
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+--------------+----------+--------------+----------+--------+
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| 1 | 1 | Laptop | 5 | 999.99 |
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| 2 | 1 | Mouse | 50 | 19.99 |
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| 3 | 1 | Keyboard | 25 | 79.99 |
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| 4 | 1 | Monitor | 15 | 299.99 |
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| 5 | 2 | Phone | 3 | 699.99 |
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| 6 | 2 | Charger | 100 | 25.99 |
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| 7 | 2 | Case | 75 | 15.99 |
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| 8 | 2 | Headphones | 20 | 149.99 |
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| 9 | 3 | Tablet | 2 | 499.99 |
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| 10 | 3 | Stylus | 80 | 29.99 |
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| 11 | 3 | Cover | 60 | 39.99 |
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| 12 | 4 | Watch | 10 | 299.99 |
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| 13 | 4 | Band | 25 | 49.99 |
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| 14 | 5 | Camera | 8 | 599.99 |
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| 15 | 5 | Lens | 12 | 199.99 |
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+--------------+----------+--------------+----------+--------+
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**Output:**
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+----------+----------------+-------------+------------------+--------------------+------------------+
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| store_id | store_name | location | most_exp_product | cheapest_product | imbalance_ratio |
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+----------+----------------+-------------+------------------+--------------------+------------------+
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| 3 | City Center | Los Angeles | Tablet | Stylus | 40.00 |
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| 1 | Downtown Tech | New York | Laptop | Mouse | 10.00 |
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| 2 | Suburb Mall | Chicago | Phone | Case | 25.00 |
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+----------+----------------+-------------+------------------+--------------------+------------------+
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**Explanation:**
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* **Downtown Tech (store\_id = 1):**
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* Most expensive product: Laptop ($999.99) with quantity 5
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* Cheapest product: Mouse ($19.99) with quantity 50
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* Inventory imbalance: 5 < 50 (expensive product has lower stock)
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* Imbalance ratio: 50 / 5 = 10.00
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* Has 4 products (≥ 3), so qualifies
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* **Suburb Mall (store\_id = 2):**
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* Most expensive product: Phone ($699.99) with quantity 3
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* Cheapest product: Case ($15.99) with quantity 75
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* Inventory imbalance: 3 < 75 (expensive product has lower stock)
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* Imbalance ratio: 75 / 3 = 25.00
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* Has 4 products (≥ 3), so qualifies
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* **City Center (store\_id = 3):**
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* Most expensive product: Tablet ($499.99) with quantity 2
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* Cheapest product: Stylus ($29.99) with quantity 80
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* Inventory imbalance: 2 < 80 (expensive product has lower stock)
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* Imbalance ratio: 80 / 2 = 40.00
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* Has 3 products (≥ 3), so qualifies
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* **Stores not included:**
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* Corner Shop (store\_id = 4): Only has 2 products (Watch, Band) - doesn't meet minimum 3 products requirement
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* Plaza Store (store\_id = 5): Only has 2 products (Camera, Lens) - doesn't meet minimum 3 products requirement
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The Results table is ordered by imbalance ratio in descending order, then by store name in ascending order
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## Solution
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```sql
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# Write your MySQL query statement below
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WITH store_product_check AS (
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SELECT
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s.store_id,
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s.store_name,
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s.location,
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COUNT(i.inventory_id) AS store_product_ct
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FROM
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stores s
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JOIN inventory i ON s.store_id = i.store_id
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GROUP BY
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s.store_id,
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s.store_name,
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s.location
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HAVING
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COUNT(i.inventory_id) >= 3
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),
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store_product_ranked AS (
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SELECT
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s.store_id,
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s.store_name,
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s.location,
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i.inventory_id,
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i.product_name,
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i.quantity,
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i.price,
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ROW_NUMBER() OVER (PARTITION BY s.store_id ORDER BY i.price ASC) AS low_price_rk,
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ROW_NUMBER() OVER (PARTITION BY s.store_id ORDER BY i.price DESC) AS high_price_rk
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FROM
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stores s
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JOIN inventory i ON s.store_id = i.store_id
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),
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high_low_price AS (
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SELECT
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spc.store_id,
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spc.store_name,
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spc.location,
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lp.product_name AS low_price_product_name,
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lp.quantity + 0.0 AS low_price_quantity,
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hp.product_name AS high_price_product_name,
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hp.quantity + 0.0 AS high_price_quantity
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FROM
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store_product_check spc
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JOIN store_product_ranked lp
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ON spc.store_id = lp.store_id AND lp.low_price_rk = 1
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JOIN store_product_ranked hp
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ON spc.store_id = hp.store_id AND hp.high_price_rk = 1
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)
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SELECT
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hlp.store_id,
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hlp.store_name,
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hlp.location,
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hlp.high_price_product_name AS most_exp_product,
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hlp.low_price_product_name AS cheapest_product,
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ROUND(hlp.low_price_quantity / hlp.high_price_quantity, 2) AS imbalance_ratio
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FROM
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high_low_price hlp
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WHERE
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hlp.high_price_quantity < hlp.low_price_quantity
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ORDER BY
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imbalance_ratio DESC,
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hlp.store_name ASC;
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```

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