Join tables to map userId to name

Problem

You are given three tables represented as 2D string arrays (each row is a record, no headers). You need to produce a mapping from userId → customerName by matching and joining information across tables.

Table 1: customers

Each row: [customerName, email, productName, quantity]

• quantity is an integer in string form.
• Assumption: each customer buys only one type of product (i.e., one productName per customer).

Table 2: orders

Each row: [userId, email, orderTotal]

• orderTotal is a number in string form (integer or decimal).
• email may be null (or an empty string, depending on how the input is represented).

Table 3: products

Each row: [productName, unitPrice]

• unitPrice is a number in string form.

Matching / Join requirement

For each order record, identify which customer placed it, then output userId -> customerName.

Use these matching rules:

1. If orders.email is present (non-null/non-empty), match the customer by email.
2. If orders.email is missing, match by using the fact that:
   • orderTotal = quantity * unitPrice for the customer’s purchased product.
   • Use customers.productName joined to products.productName to get unitPrice.

Output

Return a mapping/dictionary from userId to customerName for all orders.

Assumptions / Edge cases

• If multiple customers could match an order (e.g., same computed total), specify and implement a deterministic tie-breaker (e.g., pick lexicographically smallest name) or return all candidates—clarify with the interviewer.
• Assume inputs are well-formed aside from possible null/empty emails.

Join Tables to Map userId to Customer Name

You are given three tables represented as 2D arrays of strings (each inner list is a row, with no headers). Your task is to build a mapping from each order's userId to the corresponding customerName by joining information across the tables.

Tables:

1. customers: each row is [customerName, email, productName, quantity]

   • quantity is an integer represented as a string.
   • Each customer buys exactly one product type (one productName per customer).

2. orders: each row is [userId, email, orderTotal]

   • orderTotal is a number represented as a string (may be integer or decimal).
   • email may be missing, represented as None or an empty string "".

3. products: each row is [productName, unitPrice]

   • unitPrice is a number represented as a string.

Matching rules for each order:

1. If orders.email is present (not None and not empty) and matches one or more customer emails, match by email.
2. Otherwise, match by total price using: orderTotal = quantity * unitPrice, where unitPrice is looked up via customers.productName joined with products.productName.

Ambiguity / tie-breaker:

• If multiple customers match by email or by computed total, return the lexicographically smallest customerName among the matches.

Return a dictionary mapping userId (string) to customerName (string) for all orders.

Constraints

• 1 <= len(customers), len(orders), len(products) <= 100000
• quantity is an integer string that fits in 32-bit signed range
• unitPrice and orderTotal are valid numeric strings (integer or decimal)
• orders.email may be None or ""
• If multiple customers match, choose the lexicographically smallest customerName
• Inputs are well-formed so that each order can be matched by email or by total

Examples

Example 1

Input: ([['Alice','a@x.com','Book','2'],['Bob','b@x.com','Pen','5']], [['u1','a@x.com','20'],['u2',None,'10']], [['Book','10'],['Pen','2']])
Output: {'u1': 'Alice', 'u2': 'Bob'}

Explanation: u1 matches by email -> Alice. u2 has no email, total 10 matches Bob because 5*2=10.

Example 2

Input: ([['Amy','amy@mail','Widget','1'],['Zoe','zoe@mail','Widget','1'],['Bob','bob@mail','Gadget','2']], [['1',None,'9.99'],['2','','9.99'],['3',None,'9.990'],['4','bob@mail','10']], [['Widget','9.99'],['Gadget','5']])
Output: {'1': 'Amy', '2': 'Amy', '3': 'Amy', '4': 'Bob'}

Explanation: Orders 1-3 match by total 9.99; both Amy and Zoe match, pick lexicographically smallest -> Amy. Order 4 matches by email -> Bob.