How do you solve Find Stores with Inventory Imbalance on LeetCode?

Find Stores with Inventory Imbalance (LeetCode #3626) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying extremes (max/min) is key to solving imbalance problems.

What is the time complexity of Find Stores with Inventory Imbalance?

The optimal solution for Find Stores with Inventory Imbalance runs in O(n) time and uses O(n) space. Single pass through inventory for max/min and another for filtering, leading to linear time complexity.

What is the brute force solution for Find Stores with Inventory Imbalance?

The brute force approach for Find Stores with Inventory Imbalance is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each store's inventory for the most and least expensive products. Compare their quantities to find imbalances. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Stores with Inventory Imbalance?

Explain your thought process clearly. Consider edge cases, like stores with only one product. Practice SQL queries for efficiency.

What are common mistakes when solving Find Stores with Inventory Imbalance?

Not grouping by store when calculating max/min prices. Ignoring quantity comparisons after retrieving prices.

#3626

Find Stores with Inventory Imbalance

Medium

Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a single pass to gather the most and least expensive products for each store, then compare their quantities. This is efficient and leverages data structures for quick lookups.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to store the most and least expensive products for each store.
2Step 2: Iterate through the inventory to populate the dictionary with max and min prices and their quantities.
3Step 3: Filter stores where the quantity of the most expensive product is less than that of the cheapest.

solution.py1 lines

1SELECT store_id FROM (SELECT store_id, MAX(price) AS max_price, MIN(price) AS min_price FROM inventory GROUP BY store_id) AS prices JOIN inventory ON prices.store_id = inventory.store_id WHERE inventory.price = prices.max_price AND inventory.quantity < (SELECT quantity FROM inventory WHERE store_id = prices.store_id AND price = prices.min_price);

ℹ

Complexity note: Single pass through inventory for max/min and another for filtering, leading to linear time complexity.

1Identifying extremes (max/min) is key to solving imbalance problems.
2Using efficient data structures can significantly reduce complexity.

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