LeetCode 577: Employee Bonus Solution in Python – A Step-by-Step Guide

What Is LeetCode 577: Employee Bonus?

In LeetCode 577: Employee Bonus, you’re given two lists: employees, a list of tuples (id, name) representing employee ID and name, and bonuses, a list of tuples (id, bonus) representing employee ID and bonus amount. Your task is to return a list of [name, bonus] pairs for employees whose bonus is less than 1000, or None if no such employee exists. For example, with employees = [(1, "Alice"), (2, "Bob"), (3, "Charlie")] and bonuses = [(1, 1000), (2, 500), (3, 2000)], the result is [["Bob", 500]]. This problem builds on LeetCode 349: Intersection of Two Arrays for joining data but focuses on filtering and matching with a condition.

Problem Statement

• Input: employees (List[Tuple[int, str]])—list of (id, name); bonuses (List[Tuple[int, int]])—list of (id, bonus).
• Output: List[List[Union[str, int]]] or None—list of [name, bonus] for bonuses < 1000, or None if empty.
• Rules: Match id between lists; include only bonuses < 1000; return None if no matches.

Constraints

• 1 <= employees.length, bonuses.length <= 10⁴
• 1 <= id <= 10⁵ (unique in employees)
• 1 <= name.length <= 10
• 0 <= bonus <= 10⁶

Examples

• Input: employees = [(1,"Alice"),(2,"Bob"),(3,"Charlie")], bonuses = [(1,1000),(2,500),(3,2000)]
• Output: [["Bob",500]]
• Bob’s bonus (500) < 1000.
• Input: employees = [(1,"Alice"),(2,"Bob")], bonuses = [(1,1000),(2,2000)]
• Output: None
• No bonuses < 1000.
• Input: employees = [(1,"Alice")], bonuses = [(1,500)]
• Output: [["Alice",500]]
• Alice’s bonus (500) < 1000.

Understanding the Problem: Matching Bonuses

To solve LeetCode 577: Employee Bonus in Python, we need to join the employees and bonuses lists by id, filter for bonuses less than 1000, and return a list of [name, bonus] pairs or None if no such pairs exist, handling up to 10⁴ records efficiently. A brute-force nested loop (O(n * m)) could work but would be slow for large lists. Instead, a dictionary-based join preprocesses data for O(n + m) performance (n = employees, m = bonuses), making it fast and scalable. We’ll explore:

• Best Solution (Dictionary-Based Join): O(n + m) time, O(n) space—fast and optimal.
• Alternative Solution (Brute-Force Nested Loops): O(n * m) time, O(1) space—simple but slow.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary-Based Join

Why Dictionary Wins

The dictionary-based join solution is the best for LeetCode 577 because it efficiently matches employees to bonuses in O(n + m) time and O(n) space by using a dictionary to map employee IDs to names in one pass, then filtering bonuses in another, avoiding redundant searches. It’s like keeping an employee roster handy, quickly checking bonuses against it to spot the low earners—all in a smooth, streamlined process!

How It Works

Think of this as a bonus-matching organizer:

• Step 1: Build Employee Dictionary:
• Map id to name from employees.
• Step 2: Process Bonuses:
• Iterate bonuses, check if id exists in dictionary and bonus < 1000.
• Step 3: Collect Results:
• Build list of [name, bonus] for qualifying entries.
• Step 4: Return Result:
• Return list if non-empty, else None.
• Why It Works:
• Dictionary provides O(1) lookups.
• Single pass per list ensures efficiency.

It’s like a bonus-distribution maestro!

Step-by-Step Example

Example: employees = [(1,"Alice"),(2,"Bob"),(3,"Charlie")], bonuses = [(1,1000),(2,500),(3,2000)]

• Step 1: Build dictionary:
• id_to_name = {1: "Alice", 2: "Bob", 3: "Charlie"}.
• Step 2: Process bonuses:
• (1, 1000): 1000 ≥ 1000, skip.
• (2, 500): 500 < 1000, add ["Bob", 500].
• (3, 2000): 2000 ≥ 1000, skip.
• Step 3: Result: [["Bob", 500]].
• Result: [["Bob", 500]].

Example: employees = [(1,"Alice"),(2,"Bob")], bonuses = [(1,1000),(2,2000)]

• Step 1: id_to_name = {1: "Alice", 2: "Bob"}.
• Step 2:
• (1, 1000): Skip.
• (2, 2000): Skip.
• Step 3: Result: [].
• Step 4: Return None.
• Result: None.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom typing import List, Tuple, Optional, Union

class Solution:
    def findEmployeeBonus(self, employees: List[Tuple[int, str]], bonuses: List[Tuple[int, int]]) -> Optional[List[List[Union[str, int]]]]:
        # Step 1: Build dictionary mapping id to name
        id_to_name = {emp_id: name for emp_id, name in employees}

        # Step 2: Process bonuses and collect results
        result = []
        for emp_id, bonus in bonuses:
            if emp_id in id_to_name and bonus < 1000:
                result.append([id_to_name[emp_id], bonus])

        # Step 3: Return result or None
        return result if result else None

• Line 6: Create dictionary from employees using dict comprehension.
• Lines 9-12: Iterate bonuses, add pairs if bonus < 1000 and ID matches.
• Line 15: Return list if non-empty, else None.
• Time Complexity: O(n + m)—one pass each for employees and bonuses.
• Space Complexity: O(n)—dictionary storage.

It’s like a bonus-matching wizard!

Alternative Solution: Brute-Force Nested Loops

Why an Alternative Approach?

The brute-force nested loops approach checks each bonus against every employee to find matches, filtering for bonuses under 1000, running in O(n * m) time and O(1) space (excluding output). It’s simple but inefficient for large lists, making it a good baseline for small datasets or understanding.

How It Works

Picture this as a bonus-scanning seeker:

• Step 1: Iterate each bonus.
• Step 2: Scan employees for matching ID, check bonus < 1000.
• Step 3: Collect qualifying pairs.
• Step 4: Return list or None.

It’s like a bonus-pairing explorer!

Step-by-Step Example

Example: employees = [(1,"Alice"),(2,"Bob")], bonuses = [(1,1000),(2,500)]

• Step 1: Bonus (1, 1000):
• Check (1, "Alice"): Match, 1000 ≥ 1000, skip.
• Check (2, "Bob"): No match.
• Step 2: Bonus (2, 500):
• Check (1, "Alice"): No match.
• Check (2, "Bob"): Match, 500 < 1000, add ["Bob", 500].
• Step 3: Result: [["Bob", 500]].
• Result: [["Bob", 500]].

Code for Brute-Force Approach

Copy codefrom typing import List, Tuple, Optional, Union

class Solution:
    def findEmployeeBonus(self, employees: List[Tuple[int, str]], bonuses: List[Tuple[int, int]]) -> Optional[List[List[Union[str, int]]]]:
        # Step 1: Initialize result list
        result = []

        # Step 2: Check each bonus against employees
        for bonus_id, bonus in bonuses:
            for emp_id, name in employees:
                if emp_id == bonus_id and bonus < 1000:
                    result.append([name, bonus])
                    break  # Found match, move to next bonus

        # Step 3: Return result or None
        return result if result else None

• Line 7: Empty list for results.
• Lines 10-14: Nested loops: match IDs, filter bonus < 1000.
• Line 17: Return list or None.
• Time Complexity: O(n * m)—nested loops.
• Space Complexity: O(1)—minimal extra space.

It’s a brute-force bonus matcher!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n + m), O(n), fast.
• Cons: Extra dictionary space.
• Brute-Force (Alternative):
• Pros: O(n * m), O(1), space-efficient.
• Cons: Slower.

Dictionary wins for efficiency!

Additional Examples and Edge Cases

• Empty lists: None.
• No matches: None.
• All bonuses ≥ 1000: None.

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n + m), Space O(n).
• Brute-Force: Time O(n * m), Space O(1).

Dictionary’s the speed champ!

Key Takeaways

• Dictionary: Join finesse—learn at Python Basics!
• Brute-Force: Scan simplicity.
• Bonuses: Rewards are fun.
• Python: Dict or loops, your pick!

Final Thoughts: Distribute Those Bonuses!

LeetCode 577: Employee Bonus in Python is a practical data challenge. Dictionary-based join is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 349: Intersection of Two Arrays or LeetCode 560: Subarray Sum Equals K. Ready to reward? Head to Solve LeetCode 577 on LeetCode and match those bonuses today!