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All O`one Data Structure

Hard
50%
Updated 8/1/2025

All O`one Data Structure

What is this problem about?

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 O(1)O(1) 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.

Why is this asked in interviews?

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.

Algorithmic pattern used

The architecture of this solution is built on a Hash Map connected to a Doubly-Linked List.

  1. Each node in the doubly-linked list represents a unique frequency.
  2. Each node contains a set (or hash set) of all strings having that frequency.
  3. A separate hash map stores the association between a string and the specific node it currently resides in.
  4. When 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 O(1)O(1).

Example explanation

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".

Common mistakes candidates make

  • Not handling empty buckets: Forgetting to remove a frequency node from the linked list when the last string is moved out of it.
  • Pointer Mishaps: Losing track of the head or tail pointers during node insertions or deletions.
  • Inefficient Set Operations: Using a structure inside the bucket that isn't O(1)O(1) (like an array that requires searching).

Interview preparation tip

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 O(1)O(1). Being able to visualize the "Bucket" logic will help you communicate your solution clearly to the interviewer.

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