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Support room moves and query top-k fastest | Pinterest Interview Question

Q: Support room moves and query top-k fastest

This question evaluates understanding of dynamic data structure design and ordering semantics, specifically maintaining per-room FIFO arrival order, constant-time room counts, and efficient top-k retrieval based on hierarchical priorities.

Problem

There are R rooms labeled 0..R-1 (in increasing order), and P people labeled 0..P-1.

Initially, all people are in room 0 .
Operation move(personId) moves that person from room i to room i+1 , if i < R-1 . (If already in the last room, they stay there.)
Within each room, maintain the arrival order of people (FIFO by the time they entered that room).

Design a data structure that supports:

countInRoom(roomId) -> int
- Return how many people are currently in roomId .
- Must be O(1) time.
topKFastest(k) -> list[int]
- Return the k fastest people , where “faster” means:
  1. People in higher-numbered rooms are faster (closer to the end).
  2. For people in the same room , the one who entered that room earlier is faster.
- Return fewer than k people if k > P .

Example

R = 5, P = 5.

Operations:

move(1) , move(2) , move(1)

State:

Person 1 is in room 2
Person 2 is in room 1
Others are in room 0

So:

countInRoom(1) == 1 , countInRoom(2) == 1
topKFastest(2) returns [1, 2] (order can be [1,2] given the definition above).

Complexity expectations

Explain the time complexity of each operation and aim for efficient updates (especially avoiding scanning all people for countInRoom or topKFastest).

Problem

There are R rooms labeled 0..R-1 (in increasing order), and P people labeled 0..P-1.

Initially, all people are in room 0 .

Operation move(personId) moves that person from room i to room i+1 , if i < R-1 . (If already in the last room, they stay there.)

Within each room, maintain the arrival order of people (FIFO by the time they entered that room).

Design a data structure that supports:

countInRoom(roomId) -> int

Return how many people are currently in roomId .
Must be O(1) time.

topKFastest(k) -> list[int]

Return the k fastest people , where “faster” means:
1. People in higher-numbered rooms are faster (closer to the end).
2. For people in the same room , the one who entered that room earlier is faster.
Return fewer than k people if k > P .

Example

R = 5, P = 5.

Operations:

move(1) , move(2) , move(1)

State:

Person 1 is in room 2

Person 2 is in room 1

Others are in room 0

So:

countInRoom(1) == 1 , countInRoom(2) == 1

topKFastest(2) returns [1, 2] (order can be [1,2] given the definition above).

Support room moves and query top-k fastest

Quick Overview

Problem

Example

Complexity expectations

Comments (0)

Support room moves and query top-k fastest

Quick Overview

Problem

Example

Complexity expectations

Comments (0)