LeetCode #577 — EASY

Employee Bonus

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

Table: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| empId       | int     |
| name        | varchar |
| supervisor  | int     |
| salary      | int     |
+-------------+---------+
empId is the column with unique values for this table.
Each row of this table indicates the name and the ID of an employee in addition to their salary and the id of their manager.

Table: Bonus

+-------------+------+
| Column Name | Type |
+-------------+------+
| empId       | int  |
| bonus       | int  |
+-------------+------+
empId is the column of unique values for this table.
empId is a foreign key (reference column) to empId from the Employee table.
Each row of this table contains the id of an employee and their respective bonus.

Write a solution to report the name and bonus amount of each employee who satisfies either of the following:

The employee has a bonus less than 1000.
The employee did not get any bonus.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Employee table:
+-------+--------+------------+--------+
| empId | name   | supervisor | salary |
+-------+--------+------------+--------+
| 3     | Brad   | null       | 4000   |
| 1     | John   | 3          | 1000   |
| 2     | Dan    | 3          | 2000   |
| 4     | Thomas | 3          | 4000   |
+-------+--------+------------+--------+
Bonus table:
+-------+-------+
| empId | bonus |
+-------+-------+
| 2     | 500   |
| 4     | 2000  |
+-------+-------+
Output: 
+------+-------+
| name | bonus |
+------+-------+
| Brad | null  |
| John | null  |
| Dan  | 500   |
+------+-------+

Step 01

Brute Force Baseline

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers":{"Employee":["empId","name","supervisor","salary"],"Bonus":["empId","bonus"]},"rows":{"Employee":[[3,"Brad",null,4000],[1,"John",3,1000],[2,"Dan",3,2000],[4,"Thomas",3,4000]],"Bonus":[[2,500],[4,2000]]}}

Core Insight

What unlocks the optimal approach

If the EmpId in table Employee has no match in table Bonus, we consider that the corresponding bonus is null and null is smaller than 1000.
Inner join is the default join, we can solve the mismatching problem by using outer join.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #577: Employee Bonus
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #577: Employee Bonus
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #577: Employee Bonus
// # Write your MySQL query statement below
// SELECT name, bonus
// FROM
//     Employee
//     LEFT JOIN Bonus USING (empId)
// WHERE IFNULL(bonus, 0) < 1000;
// "#
// }

// Accepted solution for LeetCode #577: Employee Bonus
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #577: Employee Bonus
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #577: Employee Bonus
// # Write your MySQL query statement below
// SELECT name, bonus
// FROM
//     Employee
//     LEFT JOIN Bonus USING (empId)
// WHERE IFNULL(bonus, 0) < 1000;
// "#
// }

# Accepted solution for LeetCode #577: Employee Bonus
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #577: Employee Bonus
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #577: Employee Bonus
# # Write your MySQL query statement below
# SELECT name, bonus
# FROM
#     Employee
#     LEFT JOIN Bonus USING (empId)
# WHERE IFNULL(bonus, 0) < 1000;
# "#
# }

// Accepted solution for LeetCode #577: Employee Bonus
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #577: Employee Bonus
# Write your MySQL query statement below
SELECT name, bonus
FROM
    Employee
    LEFT JOIN Bonus USING (empId)
WHERE IFNULL(bonus, 0) < 1000;
"#
}

// Accepted solution for LeetCode #577: Employee Bonus
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #577: Employee Bonus
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #577: Employee Bonus
// # Write your MySQL query statement below
// SELECT name, bonus
// FROM
//     Employee
//     LEFT JOIN Bonus USING (empId)
// WHERE IFNULL(bonus, 0) < 1000;
// "#
// }

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.