LeetCode #570 — MEDIUM

Managers with at Least 5 Direct Reports

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

The Problem

Problem Statement

Table: Employee

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| department  | varchar |
| managerId   | int     |
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row of this table indicates the name of an employee, their department, and the id of their manager.
If managerId is null, then the employee does not have a manager.
No employee will be the manager of themself.

Write a solution to find managers with at least five direct reports.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Employee table:
+-----+-------+------------+-----------+
| id  | name  | department | managerId |
+-----+-------+------------+-----------+
| 101 | John  | A          | null      |
| 102 | Dan   | A          | 101       |
| 103 | James | A          | 101       |
| 104 | Amy   | A          | 101       |
| 105 | Anne  | A          | 101       |
| 106 | Ron   | B          | 101       |
+-----+-------+------------+-----------+
Output: 
+------+
| name |
+------+
| John |
+------+

Step 01

Brute Force Baseline

Problem summary: Table: Employee +-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | name | varchar | | department | varchar | | managerId | int | +-------------+---------+ id is the primary key (column with unique values) for this table. Each row of this table indicates the name of an employee, their department, and the id of their manager. If managerId is null, then the employee does not have a manager. No employee will be the manager of themself. Write a solution to find managers with at least five direct reports. Return the result table in any order. The result format is in the following example.

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": ["id", "name", "department", "managerId"]}, "rows": {"Employee": [[101, "John", "A", null],[102, "Dan", "A", 101], [103, "James", "A", 101], [104, "Amy", "A", 101], [105, "Anne", "A", 101], [106, "Ron", "B", 101]]}}

Step 02

Core Insight

What unlocks the optimal approach

Try to get all the mangerIDs that have count bigger than 5
Use the last hint's result as a table and do join with origin table at id equals to managerId
This is a very good example to show the performance of SQL code. Try to work out other solutions and you may be surprised by running time difference.
If your solution uses 'IN' function and runs more than 5 seconds, try to optimize it by using 'JOIN' instead.

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 #570: Managers with at Least 5 Direct Reports
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// import pandas as pd
// 
// 
// def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
//     # Group the employees by managerId and count the number of direct reports
//     manager_report_count = (
//         employee.groupby("managerId").size().reset_index(name="directReports")
//     )
// 
//     # Filter managers with at least five direct reports
//     result = manager_report_count[manager_report_count["directReports"] >= 5]
// 
//     # Merge with the Employee table to get the names of these managers
//     result = result.merge(
//         employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
//     )
// 
//     # Select only the 'name' column and drop the 'id' and 'directReports' columns
//     result = result[["name"]]
// 
//     return result

// Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// import pandas as pd
// 
// 
// def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
//     # Group the employees by managerId and count the number of direct reports
//     manager_report_count = (
//         employee.groupby("managerId").size().reset_index(name="directReports")
//     )
// 
//     # Filter managers with at least five direct reports
//     result = manager_report_count[manager_report_count["directReports"] >= 5]
// 
//     # Merge with the Employee table to get the names of these managers
//     result = result.merge(
//         employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
//     )
// 
//     # Select only the 'name' column and drop the 'id' and 'directReports' columns
//     result = result[["name"]]
// 
//     return result

# Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
import pandas as pd


def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
    # Group the employees by managerId and count the number of direct reports
    manager_report_count = (
        employee.groupby("managerId").size().reset_index(name="directReports")
    )

    # Filter managers with at least five direct reports
    result = manager_report_count[manager_report_count["directReports"] >= 5]

    # Merge with the Employee table to get the names of these managers
    result = result.merge(
        employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
    )

    # Select only the 'name' column and drop the 'id' and 'directReports' columns
    result = result[["name"]]

    return result

// Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// 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 #570: Managers with at Least 5 Direct Reports
// import pandas as pd
// 
// 
// def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
//     # Group the employees by managerId and count the number of direct reports
//     manager_report_count = (
//         employee.groupby("managerId").size().reset_index(name="directReports")
//     )
// 
//     # Filter managers with at least five direct reports
//     result = manager_report_count[manager_report_count["directReports"] >= 5]
// 
//     # Merge with the Employee table to get the names of these managers
//     result = result.merge(
//         employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
//     )
// 
//     # Select only the 'name' column and drop the 'id' and 'directReports' columns
//     result = result[["name"]]
// 
//     return result

// Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #570: Managers with at Least 5 Direct Reports
// import pandas as pd
// 
// 
// def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
//     # Group the employees by managerId and count the number of direct reports
//     manager_report_count = (
//         employee.groupby("managerId").size().reset_index(name="directReports")
//     )
// 
//     # Filter managers with at least five direct reports
//     result = manager_report_count[manager_report_count["directReports"] >= 5]
// 
//     # Merge with the Employee table to get the names of these managers
//     result = result.merge(
//         employee[["id", "name"]], left_on="managerId", right_on="id", how="inner"
//     )
// 
//     # Select only the 'name' column and drop the 'id' and 'directReports' columns
//     result = result[["name"]]
// 
//     return result

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.