Is All O`one Data Structure easy or hard?

All O`one Data Structure is classified as a Hard problem because it requires designing a custom data structure that guarantees O(1) operations for multiple updates and queries. The challenge comes from maintaining both minimum and maximum frequencies efficiently while keys constantly change counts.

How to solve All O`one Data Structure in O(1)?

Store each frequency in a bucket node of a doubly linked list. Maintain a hash map from key to its bucket. When incrementing or decrementing a key, move it to the neighboring bucket representing the new count, creating or removing buckets when needed. The head and tail of the list always provide the minimum and maximum keys.

All O`one Data Structure Python or Java solution

Python and Java solutions typically use dictionaries or hash maps to track key counts and map counts to sets of keys. Some implementations simulate bucket nodes using maps and linked structures. The optimal implementations still maintain O(1) average time for all operations.

What is the best approach for All O`one Data Structure?

The optimal approach uses a doubly linked list of frequency buckets combined with a hash map mapping keys to their bucket nodes. Each bucket stores keys with the same count. Increment and decrement operations move keys between adjacent buckets, keeping every operation including getMaxKey and getMinKey in O(1) time.

Is All O`one Data Structure asked at Google/Amazon/Meta?

All O`one Data Structure is a classic hard-level design problem frequently associated with top-tier companies such as Google, Meta, and Amazon. It tests understanding of hash maps, linked lists, and constant-time data structure design under strict performance constraints.

What data structure is used in All O`one Data Structure?

The core design combines a hash table with a doubly linked list of frequency buckets. The hash table provides O(1) lookup for each key, while the linked list maintains counts in sorted order so the minimum and maximum frequencies are always accessible instantly.

What is the time complexity of All O`one Data Structure?

The designed data structure must support inc, dec, getMaxKey, and getMinKey in O(1) time. Using a hash map for key lookup and a doubly linked list of frequency buckets allows constant-time insertion, deletion, and movement of keys between frequency groups.

Description Solution Video 47:29

432. All O`one Data Structure

Hard44.2% acceptance13 companies

Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.

Implement the AllOne class:

AllOne() Initializes the object of the data structure.
inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1.
dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement.
getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "".
getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "".

Note that each function must run in O(1) average time complexity.

Example 1:

Input
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
Output
[null, null, null, "hello", "hello", null, "hello", "leet"]

Explanation
AllOne allOne = new AllOne();
allOne.inc("hello");
allOne.inc("hello");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "hello"
allOne.inc("leet");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "leet"

Constraints:

1 <= key.length <= 10
key consists of lowercase English letters.
It is guaranteed that for each call to dec, key is existing in the data structure.
At most 5 * 10⁴ calls will be made to inc, dec, getMaxKey, and getMinKey.

Asked at 13 companies

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Google Amazon Microsoft Meta

+1 more

Bloomberg

Topics

47:29

Still stuck?Read 2 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class AllOne:

  def __init__(self):
      

  def inc(self, key: str) -> None:
      

  def dec(self, key: str) -> None:
      

  def getMaxKey(self) -> str:
      

  def getMinKey(self) -> str:
      


19# Your AllOne object will be instantiated and called as such:
20# obj = AllOne()
21# obj.inc(key)
22# obj.dec(key)
23# param_3 = obj.getMaxKey()
24# param_4 = obj.getMinKey()

Starter codePython 3Spaces: 4

Examples

Case 1

Input

["AllOne","inc","inc","getMaxKey","getMinKey","inc","getMaxKey","getMinKey"]
[[],["hello"],["hello"],[],[],["leet"],[],[]]

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