From ac7019a5e34a8eeb472813e9e3d8d4eb77e90e4c Mon Sep 17 00:00:00 2001
From: Jay Lee <polyglot.m@gmail.com>
Date: Wed, 15 Jun 2022 23:17:07 +0900
Subject: [PATCH] feat(leetcode/easy/1777-products-price-for-each-store.sql)

---
 .../1777-products-price-for-each-store.sql    | 56 +++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 coding-challange/leetcode/easy/1777-products-price-for-each-store/1777-products-price-for-each-store.sql

diff --git a/coding-challange/leetcode/easy/1777-products-price-for-each-store/1777-products-price-for-each-store.sql b/coding-challange/leetcode/easy/1777-products-price-for-each-store/1777-products-price-for-each-store.sql
new file mode 100644
index 00000000..40d3cc87
--- /dev/null
+++ b/coding-challange/leetcode/easy/1777-products-price-for-each-store/1777-products-price-for-each-store.sql
@@ -0,0 +1,56 @@
+# 1777-products-price-for-each-store
+# leetcode/easy/1777. Product's Price for Each Store
+# Difficulty: easy
+# URL: https://leetcode.com/problems/products-price-for-each-store/
+#
+# Input:
+# Products table:
+# +-------------+--------+-------+
+# | product_id  | store  | price |
+# +-------------+--------+-------+
+# | 0           | store1 | 95    |
+# | 0           | store3 | 105   |
+# | 0           | store2 | 100   |
+# | 1           | store1 | 70    |
+# | 1           | store3 | 80    |
+# +-------------+--------+-------+
+# Output:
+# +-------------+--------+--------+--------+
+# | product_id  | store1 | store2 | store3 |
+# +-------------+--------+--------+--------+
+# | 0           | 95     | 100    | 105    |
+# | 1           | 70     | null   | 80     |
+# +-------------+--------+--------+--------+
+# Explanation:
+# Product 0 price's are 95 for store1, 100 for store2 and, 105 for store3.
+# Product 1 price's are 70 for store1, 80 for store3 and, it's not sold in store2.
+
+
+# NOTE: best practice
+SELECT product_id,
+       MAX(CASE WHEN store = 'store1' THEN price END) AS store1,
+       MAX(CASE WHEN store = 'store2' THEN price END) AS store2,
+       MAX(CASE WHEN store = 'store3' THEN price END) AS store3
+FROM Products
+GROUP BY product_id;
+
+
+# NOTE: naive solution
+SELECT P.product_id,
+       (SELECT price
+        FROM Products
+        WHERE store = 'store1'
+          AND product_id = P.product_id)
+           AS store1,
+       (SELECT price
+        FROM Products
+        WHERE store = 'store2'
+          AND product_id = P.product_id)
+           AS store2,
+       (SELECT price
+        FROM Products
+        WHERE store = 'store3'
+          AND product_id = P.product_id)
+           AS store3
+FROM Products AS P
+GROUP BY P.product_id;