LeetCode #3626 — MEDIUM

Find Stores with Inventory Imbalance

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode
The Problem

Problem Statement

Table: stores

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| store_id    | int     |
| store_name  | varchar |
| location    | varchar |
+-------------+---------+
store_id is the unique identifier for this table.
Each row contains information about a store and its location.

Table: inventory

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| inventory_id| int     |
| store_id    | int     |
| product_name| varchar |
| quantity    | int     |
| price       | decimal |
+-------------+---------+
inventory_id is the unique identifier for this table.
Each row represents the inventory of a specific product at a specific store.

Write a solution to find stores that have inventory imbalance - stores where the most expensive product has lower stock than the cheapest product.

  • For each store, identify the most expensive product (highest price) and its quantity
  • For each store, identify the cheapest product (lowest price) and its quantity
  • A store has inventory imbalance if the most expensive product's quantity is less than the cheapest product's quantity
  • Calculate the imbalance ratio as (cheapest_quantity / most_expensive_quantity)
  • Round the imbalance ratio to 2 decimal places
  • Only include stores that have at least 3 different products

Return the result table ordered by imbalance ratio in descending order, then by store name in ascending order.

The result format is in the following example.

Example:

Input:

stores table:

+----------+----------------+-------------+
| store_id | store_name     | location    |
+----------+----------------+-------------+
| 1        | Downtown Tech  | New York    |
| 2        | Suburb Mall    | Chicago     |
| 3        | City Center    | Los Angeles |
| 4        | Corner Shop    | Miami       |
| 5        | Plaza Store    | Seattle     |
+----------+----------------+-------------+

inventory table:

+--------------+----------+--------------+----------+--------+
| inventory_id | store_id | product_name | quantity | price  |
+--------------+----------+--------------+----------+--------+
| 1            | 1        | Laptop       | 5        | 999.99 |
| 2            | 1        | Mouse        | 50       | 19.99  |
| 3            | 1        | Keyboard     | 25       | 79.99  |
| 4            | 1        | Monitor      | 15       | 299.99 |
| 5            | 2        | Phone        | 3        | 699.99 |
| 6            | 2        | Charger      | 100      | 25.99  |
| 7            | 2        | Case         | 75       | 15.99  |
| 8            | 2        | Headphones   | 20       | 149.99 |
| 9            | 3        | Tablet       | 2        | 499.99 |
| 10           | 3        | Stylus       | 80       | 29.99  |
| 11           | 3        | Cover        | 60       | 39.99  |
| 12           | 4        | Watch        | 10       | 299.99 |
| 13           | 4        | Band         | 25       | 49.99  |
| 14           | 5        | Camera       | 8        | 599.99 |
| 15           | 5        | Lens         | 12       | 199.99 |
+--------------+----------+--------------+----------+--------+

Output:

+----------+----------------+-------------+------------------+--------------------+------------------+
| store_id | store_name     | location    | most_exp_product | cheapest_product   | imbalance_ratio  |
+----------+----------------+-------------+------------------+--------------------+------------------+
| 3        | City Center    | Los Angeles | Tablet           | Stylus             | 40.00            |
| 1        | Downtown Tech  | New York    | Laptop           | Mouse              | 10.00            |
| 2        | Suburb Mall    | Chicago     | Phone            | Case               | 25.00            |
+----------+----------------+-------------+------------------+--------------------+------------------+

Explanation:

  • Downtown Tech (store_id = 1):
    • Most expensive product: Laptop ($999.99) with quantity 5
    • Cheapest product: Mouse ($19.99) with quantity 50
    • Inventory imbalance: 5 < 50 (expensive product has lower stock)
    • Imbalance ratio: 50 / 5 = 10.00
    • Has 4 products (≥ 3), so qualifies
  • Suburb Mall (store_id = 2):
    • Most expensive product: Phone ($699.99) with quantity 3
    • Cheapest product: Case ($15.99) with quantity 75
    • Inventory imbalance: 3 < 75 (expensive product has lower stock)
    • Imbalance ratio: 75 / 3 = 25.00
    • Has 4 products (≥ 3), so qualifies
  • City Center (store_id = 3):
    • Most expensive product: Tablet ($499.99) with quantity 2
    • Cheapest product: Stylus ($29.99) with quantity 80
    • Inventory imbalance: 2 < 80 (expensive product has lower stock)
    • Imbalance ratio: 80 / 2 = 40.00
    • Has 3 products (≥ 3), so qualifies
  • Stores not included:
    • Corner Shop (store_id = 4): Only has 2 products (Watch, Band) - doesn't meet minimum 3 products requirement
    • Plaza Store (store_id = 5): Only has 2 products (Camera, Lens) - doesn't meet minimum 3 products requirement

The Results table is ordered by imbalance ratio in descending order, then by store name in ascending order

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Table: stores +-------------+---------+ | Column Name | Type | +-------------+---------+ | store_id | int | | store_name | varchar | | location | varchar | +-------------+---------+ store_id is the unique identifier for this table. Each row contains information about a store and its location. Table: inventory +-------------+---------+ | Column Name | Type | +-------------+---------+ | inventory_id| int | | store_id | int | | product_name| varchar | | quantity | int | | price | decimal | +-------------+---------+ inventory_id is the unique identifier for this table. Each row represents the inventory of a specific product at a specific store. Write a solution to find stores that have inventory imbalance - stores where the most expensive product has lower stock than the cheapest product. For each store, identify the most expensive product (highest price) and its quantity For each store,

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers":{"stores":["store_id","store_name","location"],"inventory":["inventory_id","store_id","product_name","quantity","price"]},"rows":{"stores":[[1,"Downtown Tech","New York"],[2,"Suburb Mall","Chicago"],[3,"City Center","Los Angeles"],[4,"Corner Shop","Miami"],[5,"Plaza Store","Seattle"]],"inventory":[[1,1,"Laptop",5,999.99],[2,1,"Mouse",50,19.99],[3,1,"Keyboard",25,79.99],[4,1,"Monitor",15,299.99],[5,2,"Phone",3,699.99],[6,2,"Charger",100,25.99],[7,2,"Case",75,15.99],[8,2,"Headphones",20,149.99],[9,3,"Tablet",2,499.99],[10,3,"Stylus",80,29.99],[11,3,"Cover",60,39.99],[12,4,"Watch",10,299.99],[13,4,"Band",25,49.99],[14,5,"Camera",8,599.99],[15,5,"Lens",12,199.99]]}}
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Upper-end input sizes
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 #3626: Find Stores with Inventory Imbalance
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3626: Find Stores with Inventory Imbalance
// import pandas as pd
// 
// 
// def find_inventory_imbalance(
//     stores: pd.DataFrame, inventory: pd.DataFrame
// ) -> pd.DataFrame:
//     # Step 1: Identify stores with at least 3 products
//     store_counts = inventory["store_id"].value_counts()
//     valid_stores = store_counts[store_counts >= 3].index
// 
//     # Step 2: Find most expensive product for each valid store
//     # Sort by price (descending) then quantity (descending) and take first record per store
//     most_expensive = (
//         inventory[inventory["store_id"].isin(valid_stores)]
//         .sort_values(["store_id", "price", "quantity"], ascending=[True, False, False])
//         .groupby("store_id")
//         .first()
//         .reset_index()
//     )
// 
//     # Step 3: Find cheapest product for each store
//     # Sort by price (ascending) then quantity (descending) and take first record per store
//     cheapest = (
//         inventory.sort_values(
//             ["store_id", "price", "quantity"], ascending=[True, True, False]
//         )
//         .groupby("store_id")
//         .first()
//         .reset_index()
//     )
// 
//     # Step 4: Merge the two datasets on store_id
//     merged = pd.merge(
//         most_expensive, cheapest, on="store_id", suffixes=("_most", "_cheap")
//     )
// 
//     # Step 5: Filter for cases where cheapest product has higher quantity than most expensive
//     result = merged[merged["quantity_most"] < merged["quantity_cheap"]].copy()
// 
//     # Step 6: Calculate imbalance ratio (cheapest quantity / most expensive quantity)
//     result["imbalance_ratio"] = (
//         result["quantity_cheap"] / result["quantity_most"]
//     ).round(2)
// 
//     # Step 7: Merge with store information to get store names and locations
//     result = pd.merge(result, stores, on="store_id")
// 
//     # Step 8: Select and rename columns for final output
//     result = result[
//         [
//             "store_id",
//             "store_name",
//             "location",
//             "product_name_most",
//             "product_name_cheap",
//             "imbalance_ratio",
//         ]
//     ].rename(
//         columns={
//             "product_name_most": "most_exp_product",
//             "product_name_cheap": "cheapest_product",
//         }
//     )
// 
//     # Step 9: Sort by imbalance ratio (descending) then store name (ascending)
//     result = result.sort_values(
//         ["imbalance_ratio", "store_name"], ascending=[False, True]
//     ).reset_index(drop=True)
// 
//     return result
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
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.