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Problem solution · SQL

The Number of Passengers in Each Bus II

The Number of Passengers in Each Bus II: a SQL solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
63 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For The Number of Passengers in Each Bus II, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 63 lines of SQL from the credited upstream file 2153.sql.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeThe Number of Passengers in Each Bus II · SQLSQL
Use this to learn the idea, then write your own version.
WITH RECURSIVE  BusesNeighbors AS (    SELECT      bus_id,      arrival_time,      capacity,      IFNULL(LAG(arrival_time) OVER(        ORDER BY arrival_time), 0      ) AS prev_arrival_time    FROM Buses  ),  BusesMetadata AS (    SELECT      BusesNeighbors.bus_id,      BusesNeighbors.arrival_time,      BusesNeighbors.capacity,      BusesNeighbors.prev_arrival_time,      COUNT(Passengers.passenger_id) AS waiting,      ROW_NUMBER() OVER(        ORDER BY BusesNeighbors.arrival_time      ) AS `row_number`    FROM BusesNeighbors    LEFT JOIN Passengers      ON (        BusesNeighbors.prev_arrival_time < Passengers.arrival_time        AND Passengers.arrival_time <= BusesNeighbors.arrival_time)    GROUP BY 1, 2, 3  ),  Boarding AS (    SELECT      BusesMetadata.`row_number`,      BusesMetadata.bus_id,      LEAST(        BusesMetadata.capacity,        BusesMetadata.waiting      ) AS boarded,      GREATEST(        0,        BusesMetadata.waiting - BusesMetadata.capacity      ) AS not_boarded    FROM BusesMetadata    WHERE `row_number` = 1    UNION ALL    SELECT      BusesMetadata.`row_number`,      BusesMetadata.bus_id,      LEAST(        BusesMetadata.capacity,        Boarding.not_boarded + BusesMetadata.waiting      ) AS boarded,      GREATEST(        0,        Boarding.not_boarded + BusesMetadata.waiting - BusesMetadata.capacity      ) AS not_boarded    FROM BusesMetadata, Boarding    WHERE BusesMetadata.`row_number` = Boarding.`row_number` + 1  )SELECT  bus_id,  boarded AS passengers_cntFROM BoardingORDER BY 1;

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