The "All Oone Data Structure interview question" (Variation) requires implementing a specialized container for strings that tracks their frequencies. The goal is to perform inc, dec, getMaxKey, and getMinKey` in constant time. This problem is less about logic and more about choosing and connecting the right data structures. It challenges you to maintain a sorted order of frequencies while allowing immediate access to any string.
Uber, Microsoft, and Google ask the "All O`one Data Structure coding problem" to evaluate a candidate's system design intuition on a micro-scale. It requires a deep understanding of how pointers, hashes, and linked structures interact. It is a true test of whether a candidate can think beyond basic arrays and maps to build a custom, high-performance data container.
The architecture of this solution is built on a Hash Map connected to a Doubly-Linked List.
inc or dec is called, the string is moved to an adjacent node (frequency +1 or -1). If the required frequency node doesn't exist, a new one is inserted into the list in .Imagine strings "A" and "B".
inc("A"): Node(freq:1, strings:{"A"}) created.inc("B"): "B" added to Node(freq:1, strings:{"A", "B"}).inc("A"): Node(freq:2, strings:{"A"}) created. "A" moved from freq:1 node to freq:2 node.getMaxKey(): Look at the tail of the list. Tail node has freq 2, contains {"A"}. Output "A".getMinKey(): Look at the head of the list. Head node has freq 1, contains {"B"}. Output "B".This problem is a "stress test" for your pointer manipulation skills. Before your interview, practice writing a generic Doubly-Linked List that supports adding/removing nodes in . Being able to visualize the "Bucket" logic will help you communicate your solution clearly to the interviewer.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| All O`one Data Structure | Hard | Solve | |
| LFU Cache | Hard | Solve | |
| Design Authentication Manager | Medium | Solve | |
| LRU Cache | Medium | Solve | |
| Design Twitter | Medium | Solve |