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Mary is a teacher in a middle school and she has a table `seat`

storing students' names and their corresponding seat ids.

+---------+---------+ | id | student | +---------+---------+ | 1 | Abbot | | 2 | Doris | | 3 | Emerson | | 4 | Green | | 5 | Jeames | +---------+---------+For the sample input, the output is:

+---------+---------+ | id | student | +---------+---------+ | 1 | Doris | | 2 | Abbot | | 3 | Green | | 4 | Emerson | | 5 | Jeames | +---------+---------+

**Note:**

If the number of students is odd, there is no need to change the last one's seat.

b'

\n## Solution

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\n#### Approach I: Using flow control statement

\n\n\n#### Approach II: Using bit manipulation and

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`CASE`

[Accepted]**Algorithm**

For students with odd id, the new id is (id+1) after switch unless it is the last seat. And for students with even id, the new id is (id-1). In order to know how many seats in total, we can use a subquery:

\nSELECT\n COUNT(*) AS counts\nFROM\n seat\n

Then, we can use the `CASE`

statement and `MOD()`

function to alter the seat id of each student.

**MySQL**

SELECT\n (CASE\n WHEN MOD(id, 2) != 0 AND counts != id THEN id + 1\n WHEN MOD(id, 2) != 0 AND counts = id THEN id\n ELSE id - 1\n END) AS id,\n student\nFROM\n seat,\n (SELECT\n COUNT(*) AS counts\n FROM\n seat) AS seat_counts\nORDER BY id ASC;\n

`COALESCE()`

[Accepted]**Algorithm**

Bit manipulation expression `(id+1)^1-1`

can calculate the new id after switch.

SELECT id, (id+1)^1-1, student FROM seat;\n

| id | (id+1)^1-1 | student |\n|----|------------|---------|\n| 1 | 2 | Abbot |\n| 2 | 1 | Doris |\n| 3 | 4 | Emerson |\n| 4 | 3 | Green |\n| 5 | 6 | Jeames |\n

Then, we can make a temp table and join seat with this table like below.

\nSELECT\n *\nFROM\n seat s1\n LEFT JOIN\n seat s2 ON (s1.id+1)^1-1 = s2.id\nORDER BY s1.id;\n

| id | student | id | student |\n|----|---------|----|---------|\n| 1 | Abbot | 2 | Doris |\n| 2 | Doris | 1 | Abbot |\n| 3 | Emerson | 4 | Green |\n| 4 | Green | 3 | Emerson |\n| 5 | Jeames | | |\n

\n\nNote:The first two columns are from s1 and the last two are from s2.

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At last, we can output s1.id and s2.student. However, the s2.student is NULL for seat id \'5\' but s1.student is right. Thus, we we can use function `COALESCE()`

to generate the correct output for the last record.

**MySQL**

SELECT\n s1.id, COALESCE(s2.student, s1.student) AS student\nFROM\n seat s1\n LEFT JOIN\n seat s2 ON ((s1.id + 1) ^ 1) - 1 = s2.id\nORDER BY s1.id;\n

\n\nNote: This solution comes from @FANGXIAOFANG.

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