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+
+---
+
+### 📄 '04_PARTITION_BY.sql'
+
+```sql
+-- ==========================================================
+-- SQL Engineering Handbook
+--
+-- Topic:        PARTITION BY
+-- Module:       07_Window_Functions / 04_PARTITION_BY
+-- Difficulty:   Intermediate
+-- Author:       SQL Engineering Handbook Contributors
+-- Description:  Splits the result set into independent groups so a
+--               window function resets and recalculates per group,
+--               without collapsing rows the way GROUP BY does.
+-- Prerequisites: 01_ROW_NUMBER, 02_RANK, 03_DENSE_RANK, 03_Joins
+-- Dataset:      employes JOIN departments ON dept_id
+-- Learning
+-- Objectives:   1. Compute per-group rankings and counts.
+--               2. Retrieve the first row of every group.
+--               3. Build a multi-function department leaderboard.
+-- ==========================================================
+
+-- Reference data (for exploration while learning):
+SELECT * FROM departments;
+SELECT * FROM locations;
+SELECT * FROM employes;
+
+
+-- ==========================================================
+-- Q1 -- Assign Row Numbers Within Each Department
+-- ==========================================================
+-- Problem Statement:
+--   Assign a row number to each employee, restarting the sequence at
+--   the beginning of every department.
+
+SELECT
+    emp_id,
+    emp_name,
+    dept_name,
+    ROW_NUMBER() OVER (
+        PARTITION BY dept_name
+        ORDER BY emp_id
+    ) AS dept_seq
+FROM employes
+JOIN departments
+    ON employes.dept_id = departments.dept_id;
+
+-- Expected Output (shape):
+-- dept_name | emp_id | dept_seq
+-- ----------|--------|--------
+-- Sales     |  101   |   1
+-- Sales     |  105   |   2
+-- Finance   |  102   |   1     <- resets for the new partition
+--
+-- Business Use Case:
+--   Number employees within their own department for a per-team
+--   directory listing.
+
+
+-- ==========================================================
+-- Q2 -- Assign Ranks Within Each Department
+-- ==========================================================
+-- Problem Statement:
+--   Assign a rank to each employee, scoped to their department.
+
+SELECT
+    emp_id,
+    emp_name,
+    dept_name,
+    RANK() OVER (
+        PARTITION BY dept_name
+        ORDER BY emp_id
+    ) AS dept_ranks
+FROM employes
+JOIN departments
+    ON employes.dept_id = departments.dept_id;
+
+-- Business Use Case:
+--   Department-level performance leaderboards where ties should share
+--   a position within that department only.
+
+
+-- ==========================================================
+-- Q3 -- Count Employees in Each Department Using a Window Function
+-- ==========================================================
+-- Problem Statement:
+--   Show, on every employee row, the total headcount of their
+--   department -- without collapsing rows via GROUP BY.
+
+SELECT
+    emp_id,
+    emp_name,
+    dept_name,
+    COUNT(emp_id) OVER (
+        PARTITION BY dept_name
+    ) AS emp_count
+FROM employes
+JOIN departments
+    ON employes.dept_id = departments.dept_id;
+
+-- Engineering Tip:
+--   This is the key advantage of window functions over GROUP BY: you
+--   get the aggregate (emp_count) AND every individual row's detail
+--   in the same result set, with no self-join required.
+
+
+-- ==========================================================
+-- Q4 -- Show the First Employee From Every Department
+-- ==========================================================
+-- Problem Statement:
+--   Return exactly one employee per department -- the one with the
+--   lowest emp_id.
+
+WITH dept_ranks AS (
+    SELECT
+        emp_id,
+        emp_name,
+        dept_name,
+        ROW_NUMBER() OVER (
+            PARTITION BY dept_name
+            ORDER BY emp_id
+        ) AS emp_rank
+    FROM employes
+    JOIN departments
+        ON employes.dept_id = departments.dept_id
+)
+SELECT
+    emp_id,
+    emp_name,
+    dept_name,
+    emp_rank
+FROM dept_ranks
+WHERE emp_rank = 1;
+
+-- Business Use Case:
+--   The classic "top-N per group" pattern -- here N = 1. Swap
+--   `emp_rank = 1` for `emp_rank <= 3` to get the top 3 per department.
+
+
+-- ==========================================================
+-- Q5 -- Create a Department Leaderboard
+-- ==========================================================
+-- Problem Statement:
+--   Combine ROW_NUMBER(), RANK(), and DENSE_RANK() -- all partitioned
+--   by department and ordered by manager_id -- into a single
+--   leaderboard view.
+
+SELECT
+    emp_id,
+    emp_name,
+    dept_name,
+    ROW_NUMBER() OVER (
+        PARTITION BY dept_name
+        ORDER BY manager_id
+    ) AS emp_seq,
+    RANK() OVER (
+        PARTITION BY dept_name
+        ORDER BY manager_id
+    ) AS emp_ranks,
+    DENSE_RANK() OVER (
+        PARTITION BY dept_name
+        ORDER BY manager_id
+    ) AS rankings
+FROM employes
+JOIN departments
+    ON employes.dept_id = departments.dept_id;
+
+-- Business Use Case:
+--   A single query that gives report authors every ranking flavor
+--   they might need, scoped correctly to each department, ready to
+--   be filtered or exported into a dashboard.
+
+
+-- ==========================================================
+-- Summary
+-- ==========================================================
+-- PARTITION BY is the mechanism that turns any window function from a
+-- "whole table" calculation into a "per group" calculation. It is the
+-- single most reused clause in production analytics SQL.
+
+-- ==========================================================
+-- Common Mistakes
+-- ==========================================================
+-- 1. Forgetting PARTITION BY and computing a global rank by accident.
+-- 2. Joining incorrectly and duplicating rows before partitioning,
+--    which silently corrupts every downstream calculation.
+-- 3. Assuming PARTITION BY filters data -- it only scopes the window.
+
+-- ==========================================================
+-- Performance Notes
+-- ==========================================================
+-- A composite index on (dept_id, emp_id) (or the equivalent
+-- partition/order columns) lets the engine avoid a full re-sort per
+-- partition, which matters significantly once the employee table
+-- grows past a few hundred thousand rows.
+
+-- ==========================================================
+-- Interview Questions
+-- ==========================================================
+-- 1. How does PARTITION BY differ from GROUP BY?
+-- 2. How would you find the top 2 highest-paid employees per
+--    department?
+-- 3. Can you use multiple columns in PARTITION BY? Give an example.
+
+-- ==========================================================
+-- Further Reading
+-- ==========================================================
+-- https://dev.mysql.com/doc/refman/8.0/en/window-functions-usage.html