LeetCode #3611 — MEDIUM

Find Overbooked Employees

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

The Problem

Problem Statement

Table: employees

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| employee_id   | int     |
| employee_name | varchar |
| department    | varchar |
+---------------+---------+
employee_id is the unique identifier for this table.
Each row contains information about an employee and their department.

Table: meetings

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| meeting_id    | int     |
| employee_id   | int     |
| meeting_date  | date    |
| meeting_type  | varchar |
| duration_hours| decimal |
+---------------+---------+
meeting_id is the unique identifier for this table.
Each row represents a meeting attended by an employee. meeting_type can be 'Team', 'Client', or 'Training'.

Write a solution to find employees who are meeting-heavy - employees who spend more than 50% of their working time in meetings during any given week.

Assume a standard work week is 40 hours
Calculate total meeting hours per employee per week (Monday to Sunday)
An employee is meeting-heavy if their weekly meeting hours > 20 hours (50% of 40 hours)
Count how many weeks each employee was meeting-heavy
Only include employees who were meeting-heavy for at least 2 weeks

Return the result table ordered by the number of meeting-heavy weeks in descending order, then by employee name in ascending order.

The result format is in the following example.

Example:

Input:

employees table:

+-------------+----------------+-------------+
| employee_id | employee_name  | department  |
+-------------+----------------+-------------+
| 1           | Alice Johnson  | Engineering |
| 2           | Bob Smith      | Marketing   |
| 3           | Carol Davis    | Sales       |
| 4           | David Wilson   | Engineering |
| 5           | Emma Brown     | HR          |
+-------------+----------------+-------------+

meetings table:

+------------+-------------+--------------+--------------+----------------+
| meeting_id | employee_id | meeting_date | meeting_type | duration_hours |
+------------+-------------+--------------+--------------+----------------+
| 1          | 1           | 2023-06-05   | Team         | 8.0            |
| 2          | 1           | 2023-06-06   | Client       | 6.0            |
| 3          | 1           | 2023-06-07   | Training     | 7.0            |
| 4          | 1           | 2023-06-12   | Team         | 12.0           |
| 5          | 1           | 2023-06-13   | Client       | 9.0            |
| 6          | 2           | 2023-06-05   | Team         | 15.0           |
| 7          | 2           | 2023-06-06   | Client       | 8.0            |
| 8          | 2           | 2023-06-12   | Training     | 10.0           |
| 9          | 3           | 2023-06-05   | Team         | 4.0            |
| 10         | 3           | 2023-06-06   | Client       | 3.0            |
| 11         | 4           | 2023-06-05   | Team         | 25.0           |
| 12         | 4           | 2023-06-19   | Client       | 22.0           |
| 13         | 5           | 2023-06-05   | Training     | 2.0            |
+------------+-------------+--------------+--------------+----------------+

Output:

+-------------+----------------+-------------+---------------------+
| employee_id | employee_name  | department  | meeting_heavy_weeks |
+-------------+----------------+-------------+---------------------+
| 1           | Alice Johnson  | Engineering | 2                   |
| 4           | David Wilson   | Engineering | 2                   |
+-------------+----------------+-------------+---------------------+

Explanation:

Alice Johnson (employee_id = 1):
- Week of June 5-11 (2023-06-05 to 2023-06-11): 8.0 + 6.0 + 7.0 = 21.0 hours (> 20 hours)
- Week of June 12-18 (2023-06-12 to 2023-06-18): 12.0 + 9.0 = 21.0 hours (> 20 hours)
- Meeting-heavy for 2 weeks
David Wilson (employee_id = 4):
- Week of June 5-11: 25.0 hours (> 20 hours)
- Week of June 19-25: 22.0 hours (> 20 hours)
- Meeting-heavy for 2 weeks
Employees not included:
- Bob Smith (employee_id = 2): Week of June 5-11: 15.0 + 8.0 = 23.0 hours (> 20), Week of June 12-18: 10.0 hours (< 20). Only 1 meeting-heavy week
- Carol Davis (employee_id = 3): Week of June 5-11: 4.0 + 3.0 = 7.0 hours (< 20). No meeting-heavy weeks
- Emma Brown (employee_id = 5): Week of June 5-11: 2.0 hours (< 20). No meeting-heavy weeks

The result table is ordered by meeting_heavy_weeks in descending order, then by employee name in ascending order.

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":{"employees":["employee_id","employee_name","department"],"meetings":["meeting_id","employee_id","meeting_date","meeting_type","duration_hours"]},"rows":{"employees":[[1,"Alice Johnson","Engineering"],[2,"Bob Smith","Marketing"],[3,"Carol Davis","Sales"],[4,"David Wilson","Engineering"],[5,"Emma Brown","HR"]],"meetings":[[1,1,"2023-06-05","Team",8.0],[2,1,"2023-06-06","Client",6.0],[3,1,"2023-06-07","Training",7.0],[4,1,"2023-06-12","Team",12.0],[5,1,"2023-06-13","Client",9.0],[6,2,"2023-06-05","Team",15.0],[7,2,"2023-06-06","Client",8.0],[8,2,"2023-06-12","Training",10.0],[9,3,"2023-06-05","Team",4.0],[10,3,"2023-06-06","Client",3.0],[11,4,"2023-06-05","Team",25.0],[12,4,"2023-06-19","Client",22.0],[13,5,"2023-06-05","Training",2.0]]}}

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

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 #3611: Find Overbooked Employees
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3611: Find Overbooked Employees
// import pandas as pd
// 
// 
// def find_overbooked_employees(
//     employees: pd.DataFrame, meetings: pd.DataFrame
// ) -> pd.DataFrame:
//     meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
//     meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
//     meetings["week"] = meetings["meeting_date"].dt.isocalendar().week
// 
//     week_meeting_hours = (
//         meetings.groupby(["employee_id", "year", "week"], as_index=False)[
//             "duration_hours"
//         ]
//         .sum()
//         .rename(columns={"duration_hours": "hours"})
//     )
// 
//     intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]
// 
//     intensive_count = (
//         intensive_weeks.groupby("employee_id")
//         .size()
//         .reset_index(name="meeting_heavy_weeks")
//     )
// 
//     result = intensive_count.merge(employees, on="employee_id")
// 
//     result = result[result["meeting_heavy_weeks"] >= 2]
// 
//     result = result.sort_values(
//         ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
//     )
// 
//     return result[
//         ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
//     ].reset_index(drop=True)

// Accepted solution for LeetCode #3611: Find Overbooked Employees
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3611: Find Overbooked Employees
// import pandas as pd
// 
// 
// def find_overbooked_employees(
//     employees: pd.DataFrame, meetings: pd.DataFrame
// ) -> pd.DataFrame:
//     meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
//     meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
//     meetings["week"] = meetings["meeting_date"].dt.isocalendar().week
// 
//     week_meeting_hours = (
//         meetings.groupby(["employee_id", "year", "week"], as_index=False)[
//             "duration_hours"
//         ]
//         .sum()
//         .rename(columns={"duration_hours": "hours"})
//     )
// 
//     intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]
// 
//     intensive_count = (
//         intensive_weeks.groupby("employee_id")
//         .size()
//         .reset_index(name="meeting_heavy_weeks")
//     )
// 
//     result = intensive_count.merge(employees, on="employee_id")
// 
//     result = result[result["meeting_heavy_weeks"] >= 2]
// 
//     result = result.sort_values(
//         ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
//     )
// 
//     return result[
//         ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
//     ].reset_index(drop=True)

# Accepted solution for LeetCode #3611: Find Overbooked Employees
import pandas as pd


def find_overbooked_employees(
    employees: pd.DataFrame, meetings: pd.DataFrame
) -> pd.DataFrame:
    meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
    meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
    meetings["week"] = meetings["meeting_date"].dt.isocalendar().week

    week_meeting_hours = (
        meetings.groupby(["employee_id", "year", "week"], as_index=False)[
            "duration_hours"
        ]
        .sum()
        .rename(columns={"duration_hours": "hours"})
    )

    intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]

    intensive_count = (
        intensive_weeks.groupby("employee_id")
        .size()
        .reset_index(name="meeting_heavy_weeks")
    )

    result = intensive_count.merge(employees, on="employee_id")

    result = result[result["meeting_heavy_weeks"] >= 2]

    result = result.sort_values(
        ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
    )

    return result[
        ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
    ].reset_index(drop=True)

// Accepted solution for LeetCode #3611: Find Overbooked Employees
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #3611: Find Overbooked Employees
// import pandas as pd
// 
// 
// def find_overbooked_employees(
//     employees: pd.DataFrame, meetings: pd.DataFrame
// ) -> pd.DataFrame:
//     meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
//     meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
//     meetings["week"] = meetings["meeting_date"].dt.isocalendar().week
// 
//     week_meeting_hours = (
//         meetings.groupby(["employee_id", "year", "week"], as_index=False)[
//             "duration_hours"
//         ]
//         .sum()
//         .rename(columns={"duration_hours": "hours"})
//     )
// 
//     intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]
// 
//     intensive_count = (
//         intensive_weeks.groupby("employee_id")
//         .size()
//         .reset_index(name="meeting_heavy_weeks")
//     )
// 
//     result = intensive_count.merge(employees, on="employee_id")
// 
//     result = result[result["meeting_heavy_weeks"] >= 2]
// 
//     result = result.sort_values(
//         ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
//     )
// 
//     return result[
//         ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
//     ].reset_index(drop=True)

// Accepted solution for LeetCode #3611: Find Overbooked Employees
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3611: Find Overbooked Employees
// import pandas as pd
// 
// 
// def find_overbooked_employees(
//     employees: pd.DataFrame, meetings: pd.DataFrame
// ) -> pd.DataFrame:
//     meetings["meeting_date"] = pd.to_datetime(meetings["meeting_date"])
//     meetings["year"] = meetings["meeting_date"].dt.isocalendar().year
//     meetings["week"] = meetings["meeting_date"].dt.isocalendar().week
// 
//     week_meeting_hours = (
//         meetings.groupby(["employee_id", "year", "week"], as_index=False)[
//             "duration_hours"
//         ]
//         .sum()
//         .rename(columns={"duration_hours": "hours"})
//     )
// 
//     intensive_weeks = week_meeting_hours[week_meeting_hours["hours"] >= 20]
// 
//     intensive_count = (
//         intensive_weeks.groupby("employee_id")
//         .size()
//         .reset_index(name="meeting_heavy_weeks")
//     )
// 
//     result = intensive_count.merge(employees, on="employee_id")
// 
//     result = result[result["meeting_heavy_weeks"] >= 2]
// 
//     result = result.sort_values(
//         ["meeting_heavy_weeks", "employee_name"], ascending=[False, True]
//     )
// 
//     return result[
//         ["employee_id", "employee_name", "department", "meeting_heavy_weeks"]
//     ].reset_index(drop=True)

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.