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‎problems/ads-performance/README.md‎

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## [1322. Ads Performance (Easy)](https://leetcode.com/problems/ads-performance "")
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<p>Table: <code>Ads</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| ad_id | int |
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| user_id | int |
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| action | enum |
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+---------------+---------+
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(ad_id, user_id) is the primary key for this table.
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Each row of this table contains the ID of an Ad, the ID of a user and the action taken by this user regarding this Ad.
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The action column is an ENUM type of ('Clicked', 'Viewed', 'Ignored').
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</pre>
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A company is running Ads and wants to calculate the performance of each Ad.
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Performance of the Ad is measured using Click-Through Rate (CTR) where:
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[[image-blog:Leetcode: Ads Performance][https://raw.githubusercontent.com/dennyzhang/code.dennyzhang.com/master/problems/ads-performance/ctrformula.png]]
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Write an SQL query to find the ctr of each Ad.
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Round ctr to 2 decimal points. Order the result table by ctr in descending order and by ad_id in ascending order in case of a tie.
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The query result format is in the following example:
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<pre>
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Ads table:
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+-------+---------+---------+
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| ad_id | user_id | action |
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+-------+---------+---------+
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| 1 | 1 | Clicked |
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| 2 | 2 | Clicked |
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| 3 | 3 | Viewed |
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| 5 | 5 | Ignored |
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| 1 | 7 | Ignored |
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| 2 | 7 | Viewed |
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| 3 | 5 | Clicked |
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| 1 | 4 | Viewed |
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| 2 | 11 | Viewed |
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| 1 | 2 | Clicked |
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+-------+---------+---------+
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Result table:
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+-------+-------+
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| ad_id | ctr |
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+-------+-------+
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| 1 | 66.67 |
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| 3 | 50.00 |
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| 2 | 33.33 |
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| 5 | 0.00 |
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+-------+-------+
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for ad_id = 1, ctr = (2/(2+1)) * 100 = 66.67
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for ad_id = 2, ctr = (1/(1+2)) * 100 = 33.33
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for ad_id = 3, ctr = (1/(1+1)) * 100 = 50.00
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for ad_id = 5, ctr = 0.00, Note that ad_id has no clicks or views.
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Note that we don't care about Ignored Ads.
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Result table is ordered by the ctr. in case of a tie we order them by ad_id
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</pre>

‎problems/list-the-products-ordered-in-a-period/README.md‎

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## [1327. List the Products Ordered in a Period (Easy)](https://leetcode.com/problems/list-the-products-ordered-in-a-period "")
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SQL Schema
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<p>Table: <code>Products</code></p>
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<pre>
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+------------------+---------+
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| Column Name | Type |
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+------------------+---------+
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| product_id | int |
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| product_name | varchar |
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| product_category | varchar |
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+------------------+---------+
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product_id is the primary key for this table.
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This table contains data about the company's products.
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</pre>
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<p>Table: <code>Orders</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| product_id | int |
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| order_date | date |
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| unit | int |
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+---------------+---------+
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There is no primary key for this table. It may have duplicate rows.
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product_id is a foreign key to Products table.
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unit is the number of products ordered in order_date.
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</pre>
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Write an SQL query to get the names of products with greater than or equal to 100 units ordered in February 2020 and their amount.
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Return result table in any order.
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The query result format is in the following example:
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<pre>
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Products table:
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+-------------+-----------------------+------------------+
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| product_id | product_name | product_category |
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+-------------+-----------------------+------------------+
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| 1 | Leetcode Solutions | Book |
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| 2 | Jewels of Stringology | Book |
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| 3 | HP | Laptop |
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| 4 | Lenovo | Laptop |
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| 5 | Leetcode Kit | T-shirt |
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+-------------+-----------------------+------------------+
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Orders table:
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+--------------+--------------+----------+
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| product_id | order_date | unit |
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+--------------+--------------+----------+
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| 1 | 2020年02月05日 | 60 |
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| 1 | 2020年02月10日 | 70 |
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| 2 | 2020年01月18日 | 30 |
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| 2 | 2020年02月11日 | 80 |
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| 3 | 2020年02月17日 | 2 |
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| 3 | 2020年02月24日 | 3 |
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| 4 | 2020年03月01日 | 20 |
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| 4 | 2020年03月04日 | 30 |
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| 4 | 2020年03月04日 | 60 |
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| 5 | 2020年02月25日 | 50 |
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| 5 | 2020年02月27日 | 50 |
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| 5 | 2020年03月01日 | 50 |
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+--------------+--------------+----------+
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Result table:
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+--------------------+---------+
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| product_name | unit |
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+--------------------+---------+
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| Leetcode Solutions | 130 |
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| Leetcode Kit | 100 |
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+--------------------+---------+
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Products with product_id = 1 is ordered in February a total of (60 + 70) = 130.
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Products with product_id = 2 is ordered in February a total of 80.
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Products with product_id = 3 is ordered in February a total of (2 + 3) = 5.
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Products with product_id = 4 was not ordered in February 2020.
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Products with product_id = 5 is ordered in February a total of (50 + 50) = 100.
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</pre>

‎problems/number-of-transactions-per-visit/README.md‎

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## [1336. Number of Transactions per Visit (Medium)](https://leetcode.com/problems/number-of-transactions-per-visit "")
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<p>Table: <code>Visits</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| user_id | int |
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| visit_date | date |
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+---------------+---------+
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(user_id, visit_date) is the primary key for this table.
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Each row of this table indicates that user_id has visited the bank in visit_date.
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</pre>
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<p>Table: <code>Transactions</code></p>
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<pre>
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+------------------+---------+
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| Column Name | Type |
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+------------------+---------+
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| user_id | int |
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| transaction_date | date |
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| amount | int |
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+------------------+---------+
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There is no primary key for this table, it may contain duplicates.
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Each row of this table indicates that user_id has done a transaction of amount in transaction_date.
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It is guaranteed that the user has visited the bank in the transaction_date.(i.e The Visits table contains (user_id, transaction_date) in one row)
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</pre>
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Write an SQL query to find how many users visited the bank and didn't do any transactions, how many visited the bank and did one transaction and so on.
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The result table will contain two columns transactions_count which is the number of transactions done in one visit and visits_count which is the corresponding number of users who did transactions_count in one visit to the bank. transactions_count should take all values from 0 to max(transactions_count) done by one or more users.
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Order the result table by transactions_count.
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The query result format is in the following example:
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<pre>
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Visits table:
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+---------+------------+
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| user_id | visit_date |
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+---------+------------+
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| 1 | 2020年01月01日 |
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| 2 | 2020年01月02日 |
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| 12 | 2020年01月01日 |
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| 19 | 2020年01月03日 |
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| 1 | 2020年01月02日 |
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| 2 | 2020年01月03日 |
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| 1 | 2020年01月04日 |
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| 7 | 2020年01月11日 |
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| 9 | 2020年01月25日 |
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| 8 | 2020年01月28日 |
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+---------+------------+
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Transactions table:
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+---------+------------------+--------+
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| user_id | transaction_date | amount |
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+---------+------------------+--------+
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| 1 | 2020年01月02日 | 120 |
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| 2 | 2020年01月03日 | 22 |
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| 7 | 2020年01月11日 | 232 |
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| 1 | 2020年01月04日 | 7 |
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| 9 | 2020年01月25日 | 33 |
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| 9 | 2020年01月25日 | 66 |
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| 8 | 2020年01月28日 | 1 |
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| 9 | 2020年01月25日 | 99 |
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+---------+------------------+--------+
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Result table:
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+--------------------+--------------+
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| transactions_count | visits_count |
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+--------------------+--------------+
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| 0 | 4 |
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| 1 | 5 |
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| 2 | 0 |
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| 3 | 1 |
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+--------------------+--------------+
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Users 1, 2, 12 and 19 visited the bank in 2020年01月01日, 2020年01月02日, 2020年01月01日 and 2020年01月03日 respectively, and didn't do any transactions.
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So we have visits_count = 4 for transactions_count = 0.
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Users 2, 7 and 8 visited the bank in 2020年01月03日, 2020年01月11日 and 2020年01月28日 respectively, and did one transaction.
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User 1 Also visited the bank in 2020年01月02日 and 2020年01月04日 and did one transaction each day.
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So we have total visits_count = 5 for transactions_count = 1.
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For transactions_count = 2 we don't have any users who visited the bank and did two transactions.
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For transactions_count = 3 we have user 9 who visited the bank in 2020年01月25日 and did three transactions.
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Note that we stopped at transactions_count = 3 as this is the maximum number of transactions done by one user in one visit to the bank.
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</pre>

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