LeetCode #262 — HARD

Trips and Users

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

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The Problem

Problem Statement

Table: Trips

+-------------+----------+
| Column Name | Type     |
+-------------+----------+
| id          | int      |
| client_id   | int      |
| driver_id   | int      |
| city_id     | int      |
| status      | enum     |
| request_at  | varchar  |     
+-------------+----------+
id is the primary key (column with unique values) for this table.
The table holds all taxi trips. Each trip has a unique id, while client_id and driver_id are foreign keys to the users_id at the Users table.
Status is an ENUM (category) type of ('completed', 'cancelled_by_driver', 'cancelled_by_client').

Table: Users

+-------------+----------+
| Column Name | Type     |
+-------------+----------+
| users_id    | int      |
| banned      | enum     |
| role        | enum     |
+-------------+----------+
users_id is the primary key (column with unique values) for this table.
The table holds all users. Each user has a unique users_id, and role is an ENUM type of ('client', 'driver', 'partner').
banned is an ENUM (category) type of ('Yes', 'No').

The cancellation rate is computed by dividing the number of canceled (by client or driver) requests with unbanned users by the total number of requests with unbanned users on that day.

Write a solution to find the cancellation rate of requests with unbanned users (both client and driver must not be banned) each day between "2013-10-01" and "2013-10-03" with at least one trip. Round Cancellation Rate to two decimal points.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input: 
Trips table:
+----+-----------+-----------+---------+---------------------+------------+
| id | client_id | driver_id | city_id | status              | request_at |
+----+-----------+-----------+---------+---------------------+------------+
| 1  | 1         | 10        | 1       | completed           | 2013-10-01 |
| 2  | 2         | 11        | 1       | cancelled_by_driver | 2013-10-01 |
| 3  | 3         | 12        | 6       | completed           | 2013-10-01 |
| 4  | 4         | 13        | 6       | cancelled_by_client | 2013-10-01 |
| 5  | 1         | 10        | 1       | completed           | 2013-10-02 |
| 6  | 2         | 11        | 6       | completed           | 2013-10-02 |
| 7  | 3         | 12        | 6       | completed           | 2013-10-02 |
| 8  | 2         | 12        | 12      | completed           | 2013-10-03 |
| 9  | 3         | 10        | 12      | completed           | 2013-10-03 |
| 10 | 4         | 13        | 12      | cancelled_by_driver | 2013-10-03 |
+----+-----------+-----------+---------+---------------------+------------+
Users table:
+----------+--------+--------+
| users_id | banned | role   |
+----------+--------+--------+
| 1        | No     | client |
| 2        | Yes    | client |
| 3        | No     | client |
| 4        | No     | client |
| 10       | No     | driver |
| 11       | No     | driver |
| 12       | No     | driver |
| 13       | No     | driver |
+----------+--------+--------+
Output: 
+------------+-------------------+
| Day        | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33              |
| 2013-10-02 | 0.00              |
| 2013-10-03 | 0.50              |
+------------+-------------------+
Explanation: 
On 2013-10-01:
  - There were 4 requests in total, 2 of which were canceled.
  - However, the request with Id=2 was made by a banned client (User_Id=2), so it is ignored in the calculation.
  - Hence there are 3 unbanned requests in total, 1 of which was canceled.
  - The Cancellation Rate is (1 / 3) = 0.33
On 2013-10-02:
  - There were 3 requests in total, 0 of which were canceled.
  - The request with Id=6 was made by a banned client, so it is ignored.
  - Hence there are 2 unbanned requests in total, 0 of which were canceled.
  - The Cancellation Rate is (0 / 2) = 0.00
On 2013-10-03:
  - There were 3 requests in total, 1 of which was canceled.
  - The request with Id=8 was made by a banned client, so it is ignored.
  - Hence there are 2 unbanned request in total, 1 of which were canceled.
  - The Cancellation Rate is (1 / 2) = 0.50

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Table: Trips +-------------+----------+ | Column Name | Type | +-------------+----------+ | id | int | | client_id | int | | driver_id | int | | city_id | int | | status | enum | | request_at | varchar | +-------------+----------+ id is the primary key (column with unique values) for this table. The table holds all taxi trips. Each trip has a unique id, while client_id and driver_id are foreign keys to the users_id at the Users table. Status is an ENUM (category) type of ('completed', 'cancelled_by_driver', 'cancelled_by_client'). Table: Users +-------------+----------+ | Column Name | Type | +-------------+----------+ | users_id | int | | banned | enum | | role | enum | +-------------+----------+ users_id is the primary key (column with unique values) for this table. The table holds all users. Each user has a unique users_id, and role is an ENUM type of ('client', 'driver', 'partner').

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers": {"Trips": ["id", "client_id", "driver_id", "city_id", "status", "request_at"], "Users": ["users_id", "banned", "role"]}, "rows": {"Trips": [["1", "1", "10", "1", "completed", "2013-10-01"], ["2", "2", "11", "1", "cancelled_by_driver", "2013-10-01"], ["3", "3", "12", "6", "completed", "2013-10-01"], ["4", "4", "13", "6", "cancelled_by_client", "2013-10-01"], ["5", "1", "10", "1", "completed", "2013-10-02"], ["6", "2", "11", "6", "completed", "2013-10-02"], ["7", "3", "12", "6", "completed", "2013-10-02"], ["8", "2", "12", "12", "completed", "2013-10-03"], ["9", "3", "10", "12", "completed", "2013-10-03"], ["10", "4", "13", "12", "cancelled_by_driver", "2013-10-03"]], "Users": [["1", "No", "client"], ["2", "Yes", "client"], ["3", "No", "client"], ["4", "No", "client"], ["10", "No", "driver"], ["11", "No", "driver"], ["12", "No", "driver"], ["13", "No", "driver"]]}}

Related Problems

  • Hopper Company Queries I (hopper-company-queries-i)
  • Hopper Company Queries II (hopper-company-queries-ii)
  • Hopper Company Queries III (hopper-company-queries-iii)
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

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #262: Trips and Users
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #262: Trips and Users
// import pandas as pd
// 
// 
// def trips_and_users(trips: pd.DataFrame, users: pd.DataFrame) -> pd.DataFrame:
//     # 1) temporal filtering
//     trips = trips[trips["request_at"].between("2013-10-01", "2013-10-03")].rename(
//         columns={"request_at": "Day"}
//     )
// 
//     # 2) filtering based not banned
//     # 2.1) mappning the column 'banned' to `client_id` and `driver_id`
//     df_client = (
//         pd.merge(trips, users, left_on="client_id", right_on="users_id", how="left")
//         .drop(["users_id", "role"], axis=1)
//         .rename(columns={"banned": "banned_client"})
//     )
//     df_driver = (
//         pd.merge(trips, users, left_on="driver_id", right_on="users_id", how="left")
//         .drop(["users_id", "role"], axis=1)
//         .rename(columns={"banned": "banned_driver"})
//     )
//     df = pd.merge(
//         df_client,
//         df_driver,
//         left_on=["id", "driver_id", "client_id", "city_id", "status", "Day"],
//         right_on=["id", "driver_id", "client_id", "city_id", "status", "Day"],
//         how="left",
//     )
//     # 2.2) filtering based on not banned
//     df = df[(df["banned_client"] == "No") & (df["banned_driver"] == "No")]
// 
//     # 3) counting the cancelled and total trips per day
//     df["status_cancelled"] = df["status"].str.contains("cancelled")
//     df = df[["Day", "status_cancelled"]]
//     df = df.groupby("Day").agg(
//         {"status_cancelled": [("total_cancelled", "sum"), ("total", "count")]}
//     )
//     df.columns = df.columns.droplevel()
//     df = df.reset_index()
// 
//     # 4) calculating the ratio
//     df["Cancellation Rate"] = (df["total_cancelled"] / df["total"]).round(2)
//     return df[["Day", "Cancellation Rate"]]
Step 06

Interactive Study Demo

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

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.