From a17f3aeaad981582bea4ba590e75f386a63a4799 Mon Sep 17 00:00:00 2001
From: pallasivasai <psairabel143@gmail.com>
Date: Sun, 14 Jul 2024 07:38:21 +0530
Subject: [PATCH 1/2] Solution for a problem 183 is added

---
 .../183-Customers-Who-Never-Order.md          | 96 +++++++++++++++++++
 1 file changed, 96 insertions(+)
 create mode 100644 dsa-solutions/lc-solutions/0100-0199/183-Customers-Who-Never-Order.md

diff --git a/dsa-solutions/lc-solutions/0100-0199/183-Customers-Who-Never-Order.md b/dsa-solutions/lc-solutions/0100-0199/183-Customers-Who-Never-Order.md
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+---
+id: customers-who-never-order
+title: Customers Who Never Order
+sidebar_label: 0183. Customers Who Never Order
+tags:
+- SQL
+- Database
+description: "Solution to Leetcode 183. Customers Who Never Order"
+---
+
+## Problem Description
+
+Given two tables, `Customers` and `Orders`, write a query to find all customers who never ordered anything.
+
+**Table: Customers**
+
+| Column Name | Type    |
+| ----------- | ------- |
+| id          | int     |
+| name        | varchar |
+id is the primary key for this table.
+
+**Table: Orders**
+
+| Column Name | Type |
+| ----------- | ---- |
+| id          | int  |
+| customerId  | int  |
+id is the primary key for this table. `customerId` is a foreign key referencing the `id` column in the `Customers` table.
+
+### Examples
+
+**Example 1:**
+
+**Input:**
+
+**Customers table:**
+
+| id  | name  |
+| --- | ----- |
+| 1   | Joe   |
+| 2   | Henry |
+| 3   | Sam   |
+| 4   | Max   |
+
+**Orders table:**
+
+| id  | customerId |
+| --- | ---------- |
+| 1   | 3          |
+| 2   | 1          |
+
+**Output:**
+
+| Customers |
+| --------- |
+| Henry     |
+| Max       |
+
+### Approach
+
+1. Perform a `LEFT JOIN` between the `Customers` table and the `Orders` table on the `id` and `customerId` columns.
+2. Use the `WHERE` clause to filter out the customers who have `NULL` in the `Orders` table, which indicates that they never placed an order.
+
+### Solution
+
+#### SQL
+
+```sql
+-- SQL solution to find customers who never order anything
+
+SELECT
+    name AS Customers
+FROM
+    Customers
+LEFT JOIN
+    Orders
+ON
+    Customers.id = Orders.customerId
+WHERE
+    Orders.customerId IS NULL;
+```
+
+### Explanation
+
+- **LEFT JOIN**: Joins the `Customers` table with the `Orders` table. This join includes all customers, even those who have not placed any orders.
+- **WHERE Orders.customerId IS NULL**: Filters the result to include only those customers who do not have a corresponding entry in the `Orders` table.
+
+### Complexity Analysis
+
+- **Time Complexity**: O(n), where n is the number of rows in the `Customers` table.
+- **Space Complexity**: O(n), where n is the number of rows in the `Customers` table.
+
+### References
+
+- **LeetCode Problem**: Customers Who Never Order
\ No newline at end of file

From e5a71495996b933c0f5428c4e2760334e26a21b4 Mon Sep 17 00:00:00 2001
From: pallasivasai <psairabel143@gmail.com>
Date: Sun, 14 Jul 2024 08:01:22 +0530
Subject: [PATCH 2/2] adding Solution for a problem 182

---
 .../0100-0199/0182-Duplicate-Emails.md        | 73 +++++++++++++++++++
 1 file changed, 73 insertions(+)
 create mode 100644 dsa-solutions/lc-solutions/0100-0199/0182-Duplicate-Emails.md

diff --git a/dsa-solutions/lc-solutions/0100-0199/0182-Duplicate-Emails.md b/dsa-solutions/lc-solutions/0100-0199/0182-Duplicate-Emails.md
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+++ b/dsa-solutions/lc-solutions/0100-0199/0182-Duplicate-Emails.md
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+---
+id: duplicate-emails
+title: Duplicate Emails
+sidebar_label: 0183. Duplicate Emails
+tags:
+- SQL
+- Database
+description: "Solution to Leetcode 183. Duplicate Emails"
+---
+
+## Problem Description
+
+Given a table `Person`, write a query to find all duplicate emails.
+
+**Table: Person**
+
+| Column Name | Type    |
+| ----------- | ------- |
+| id          | int     |
+| email       | varchar |
+id is the primary key for this table. Each row of this table contains an email. The emails will not contain uppercase letters.
+
+### Examples
+
+**Example 1:**
+
+**Input:**
+
+**Person table:**
+
+| id  | email   |
+| --- | ------- |
+| 1   | a@b.com |
+| 2   | c@d.com |
+| 3   | a@b.com |
+
+**Output:**
+
+| Email   |
+| ------- |
+| a@b.com |
+
+### Approach
+
+1. Group the rows in the `Person` table by the `email` column.
+2. Use the `HAVING` clause to filter out the groups that have a count greater than 1, indicating duplicate emails.
+
+### Solution
+
+#### SQL
+
+```sql
+-- SQL solution to find duplicate emails
+
+SELECT email
+FROM Person
+GROUP BY email
+HAVING COUNT(*) > 1;
+```
+
+### Explanation
+
+- **GROUP BY email**: Groups the records by the `email` column.
+- **HAVING COUNT(*) > 1**: Filters the groups to include only those with more than one occurrence, indicating that the email is duplicated.
+
+### Complexity Analysis
+
+- **Time Complexity**: O(n), where n is the number of rows in the `Person` table.
+- **Space Complexity**: O(n), where n is the number of rows in the `Person` table.
+
+### References
+
+- **LeetCode Problem**: Duplicate Emails
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