# Queries Quality And Percentage Problem

## Description

LeetCode Problem 1211.

Table: Queries

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+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| query_name  | varchar |
| result      | varchar |
| position    | int     |
| rating      | int     |
+-------------+---------+
There is no primary key for this table, it may have duplicate rows.
This table contains information collected from some queries on a database.
The position column has a value from 1 to 500.
The rating column has a value from 1 to 5. Query with rating less than 3 is a poor query.
``````

We define query quality as: The average of the ratio between query rating and its position.

We also define poor query percentage as: The percentage of all queries with rating less than 3.

Write an SQL query to find each query_name, the quality and poor_query_percentage.

Both quality and poor_query_percentage should be rounded to 2 decimal places.

The query result format is in the following example:

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Queries table:
+------------+-------------------+----------+--------+
| query_name | result            | position | rating |
+------------+-------------------+----------+--------+
| Dog        | Golden Retriever  | 1        | 5      |
| Dog        | German Shepherd   | 2        | 5      |
| Dog        | Mule              | 200      | 1      |
| Cat        | Shirazi           | 5        | 2      |
| Cat        | Siamese           | 3        | 3      |
| Cat        | Sphynx            | 7        | 4      |
+------------+-------------------+----------+--------+

Result table:
+------------+---------+-----------------------+
| query_name | quality | poor_query_percentage |
+------------+---------+-----------------------+
| Dog        | 2.50    | 33.33                 |
| Cat        | 0.66    | 33.33                 |
+------------+---------+-----------------------+

Dog queries quality is ((5 / 1) + (5 / 2) + (1 / 200)) / 3 = 2.50
Dog queries poor_ query_percentage is (1 / 3) * 100 = 33.33

Cat queries quality equals ((2 / 5) + (3 / 3) + (4 / 7)) / 3 = 0.66
Cat queries poor_ query_percentage is (1 / 3) * 100 = 33.33
``````

## MySQL Solution

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select query_name,
round(sum(rating/position)/count(query_name), 2) as quality,
round(100*sum(if(rating < 3, 1, 0))/count(query_name), 2) as poor_query_percentage
from Queries
group by query_name
``````