## 579. Find Cumulative Salary of an Employee

The Employee table holds the salary information in a year.

Write a SQL to get the cumulative sum of an employee's salary over a period of 3 months but exclude the most recent month.

The result should be displayed by 'Id' ascending, and then by 'Month' descending.

Example
Input

```| Id | Month | Salary |
|----|-------|--------|
| 1  | 1     | 20     |
| 2  | 1     | 20     |
| 1  | 2     | 30     |
| 2  | 2     | 30     |
| 3  | 2     | 40     |
| 1  | 3     | 40     |
| 3  | 3     | 60     |
| 1  | 4     | 60     |
| 3  | 4     | 70     |
```
Output
```| Id | Month | Salary |
|----|-------|--------|
| 1  | 3     | 90     |
| 1  | 2     | 50     |
| 1  | 1     | 20     |
| 2  | 1     | 20     |
| 3  | 3     | 100    |
| 3  | 2     | 40     |
```

Explanation

Employee '1' has 3 salary records for the following 3 months except the most recent month '4': salary 40 for month '3', 30 for month '2' and 20 for month '1'
So the cumulative sum of salary of this employee over 3 months is 90(40+30+20), 50(30+20) and 20 respectively.

```| Id | Month | Salary |
|----|-------|--------|
| 1  | 3     | 90     |
| 1  | 2     | 50     |
| 1  | 1     | 20     |
```
Employee '2' only has one salary record (month '1') except its most recent month '2'.
```| Id | Month | Salary |
|----|-------|--------|
| 2  | 1     | 20     |
```

Employ '3' has two salary records except its most recent pay month '4': month '3' with 60 and month '2' with 40. So the cumulative salary is as following.
```| Id | Month | Salary |
|----|-------|--------|
| 3  | 3     | 100    |
| 3  | 2     | 40     |
```

b'
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## Solution

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#### Approach: Using `OUTER JOIN` and temporary tables [Accepted]

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Intuition

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Solve this issue by two steps. The first one is to get the cumulative sum of an employee\'s salary over a period of 3 months, and then exclude the most recent month from the result.

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Algorithm

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If you feel hard to work out how to get the cumulative sum of an employee\'s salary over a period of 3 months, think about 2 months as a start. By joining this Employee table with itself, you can get salary information for one more month.

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`SELECT *\nFROM\n    Employee E1\n        LEFT JOIN\n    Employee E2 ON (E2.id = E1.id\n        AND E2.month = E1.month - 1)\nORDER BY E1.id ASC , E1. month DESC\n`
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IdMonthSalaryIdMonthSalary
14601340
13401230
12301120
1120
22302120
2120
34703360
33603240
3240
>Note:
> - The blank value in the output is actually `NULL` in the database.
> - The first three columns are from E1, and the rest ones are from E2.
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Then we can add the salary to get the cumulative sum for 2 months.

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`SELECT\n    E1.id,\n    E1.month,\n    (IFNULL(E1.salary, 0) + IFNULL(E2.salary, 0)) AS Salary\nFROM\n    Employee E1\n        LEFT JOIN\n    Employee E2 ON (E2.id = E1.id\n        AND E2.month = E1.month - 1)\nORDER BY E1.id ASC , E1.month DESC\n`
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`| id | month | Salary |\n|----|-------|--------|\n| 1  | 4     | 100    |\n| 1  | 3     | 70     |\n| 1  | 2     | 50     |\n| 1  | 1     | 20     |\n| 2  | 2     | 50     |\n| 2  | 1     | 20     |\n| 3  | 4     | 130    |\n| 3  | 3     | 100    |\n| 3  | 2     | 40     |\n`
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Similarly, you can join this table one more time to get the cumulative sum for 3 months.

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`SELECT\n    E1.id,\n    E1.month,\n    (IFNULL(E1.salary, 0) + IFNULL(E2.salary, 0) + IFNULL(E3.salary, 0)) AS Salary\nFROM\n    Employee E1\n        LEFT JOIN\n    Employee E2 ON (E2.id = E1.id\n        AND E2.month = E1.month - 1)\n        LEFT JOIN\n    Employee E3 ON (E3.id = E1.id\n        AND E3.month = E1.month - 2)\nORDER BY E1.id ASC , E1.month DESC\n;\n`
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`| id | month | Salary |\n|----|-------|--------|\n| 1  | 4     | 130    |\n| 1  | 3     | 90     |\n| 1  | 2     | 50     |\n| 1  | 1     | 20     |\n| 2  | 2     | 50     |\n| 2  | 1     | 20     |\n| 3  | 4     | 170    |\n| 3  | 3     | 100    |\n| 3  | 2     | 40     |\n`
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In addition, we have to exclude the most recent month as required. If we have a temp table including every id and most recent month like below, then we can easily opt out these months by join it with the above table.

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`| id | month |\n|----|-------|\n| 1  | 4     |\n| 2  | 2     |\n| 3  | 4     |\n`
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Here is the code to generate this table.

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`SELECT\n    id, MAX(month) AS month\nFROM\n    Employee\nGROUP BY id\nHAVING COUNT(*) > 1\n;\n`
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At last, we can join them together and get the desired cumulative sum of an employee\'s salary over a period of 3 months excluding the most recent one.

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MySQL

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`SELECT\n    E1.id,\n    E1.month,\n    (IFNULL(E1.salary, 0) + IFNULL(E2.salary, 0) + IFNULL(E3.salary, 0)) AS Salary\nFROM\n    (SELECT\n        id, MAX(month) AS month\n    FROM\n        Employee\n    GROUP BY id\n    HAVING COUNT(*) > 1) AS maxmonth\n        LEFT JOIN\n    Employee E1 ON (maxmonth.id = E1.id\n        AND maxmonth.month > E1.month)\n        LEFT JOIN\n    Employee E2 ON (E2.id = E1.id\n        AND E2.month = E1.month - 1)\n        LEFT JOIN\n    Employee E3 ON (E3.id = E1.id\n        AND E3.month = E1.month - 2)\nORDER BY id ASC , month DESC\n;\n`
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idmonthSalary
1390
1250
1120
2120
33100
3240
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Note: Thank @xiaxin for providing this elegant solution.

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