LeetCode #626 — MEDIUM

Exchange Seats

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Table: Seat

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| student     | varchar |
+-------------+---------+
id is the primary key (unique value) column for this table.
Each row of this table indicates the name and the ID of a student.
The ID sequence always starts from 1 and increments continuously.

Write a solution to swap the seat id of every two consecutive students. If the number of students is odd, the id of the last student is not swapped.

Return the result table ordered by id in ascending order.

The result format is in the following example.

Example 1:

Input: 
Seat table:
+----+---------+
| id | student |
+----+---------+
| 1  | Abbot   |
| 2  | Doris   |
| 3  | Emerson |
| 4  | Green   |
| 5  | Jeames  |
+----+---------+
Output: 
+----+---------+
| id | student |
+----+---------+
| 1  | Doris   |
| 2  | Abbot   |
| 3  | Green   |
| 4  | Emerson |
| 5  | Jeames  |
+----+---------+
Explanation: 
Note that if the number of students is odd, there is no need to change the last one's seat.

Step 01

Brute Force Baseline

Problem summary: Table: Seat +-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | student | varchar | +-------------+---------+ id is the primary key (unique value) column for this table. Each row of this table indicates the name and the ID of a student. The ID sequence always starts from 1 and increments continuously. Write a solution to swap the seat id of every two consecutive students. If the number of students is odd, the id of the last student is not swapped. Return the result table ordered by id in ascending order. The result format is in the following example.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers": {"Seat": ["id","student"]}, "rows": {"Seat": [[1,"Abbot"],[2,"Doris"],[3,"Emerson"],[4,"Green"],[5,"Jeames"]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #626: Exchange Seats
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #626: Exchange Seats
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #626: Exchange Seats
// # Write your MySQL query statement below
// SELECT s1.id, COALESCE(s2.student, s1.student) AS student
// FROM
//     Seat AS s1
//     LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
// ORDER BY 1;
// "#
// }

// Accepted solution for LeetCode #626: Exchange Seats
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #626: Exchange Seats
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #626: Exchange Seats
// # Write your MySQL query statement below
// SELECT s1.id, COALESCE(s2.student, s1.student) AS student
// FROM
//     Seat AS s1
//     LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
// ORDER BY 1;
// "#
// }

# Accepted solution for LeetCode #626: Exchange Seats
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #626: Exchange Seats
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #626: Exchange Seats
# # Write your MySQL query statement below
# SELECT s1.id, COALESCE(s2.student, s1.student) AS student
# FROM
#     Seat AS s1
#     LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
# ORDER BY 1;
# "#
# }

// Accepted solution for LeetCode #626: Exchange Seats
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #626: Exchange Seats
# Write your MySQL query statement below
SELECT s1.id, COALESCE(s2.student, s1.student) AS student
FROM
    Seat AS s1
    LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
ORDER BY 1;
"#
}

// Accepted solution for LeetCode #626: Exchange Seats
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #626: Exchange Seats
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #626: Exchange Seats
// # Write your MySQL query statement below
// SELECT s1.id, COALESCE(s2.student, s1.student) AS student
// FROM
//     Seat AS s1
//     LEFT JOIN Seat AS s2 ON (s1.id + 1) ^ 1 - 1 = s2.id
// ORDER BY 1;
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.