How do you solve All O`one Data Structure on LeetCode?

All O`one Data Structure (LeetCode #432) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Using a linked list allows for efficient insertions and deletions while maintaining order.

What is the brute force solution for All O`one Data Structure?

The brute force approach for All O`one Data Structure is Brute Force, with O(n²) time complexity and O(n) space complexity. In this approach, we maintain a simple dictionary to count occurrences of each string. To find the minimum and maximum keys, we traverse the entire dictionary each time, which is inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve All O`one Data Structure?

All O`one Data Structure uses the following concepts: Hash Table, Linked List, Design, Doubly-Linked List. The recommended patterns to study are: Hash Map, Doubly Linked List.

What are the interview tips for All O`one Data Structure?

Clearly explain your thought process and how you plan to manage the data structure. Consider edge cases, such as removing keys and updating counts. Practice implementing both the brute force and optimal solutions to understand the differences.

What are common mistakes when solving All O`one Data Structure?

Not properly managing the linked list when incrementing or decrementing counts. Failing to update the hash map correctly, leading to incorrect counts.

#432

All O`one Data Structure

Hard

Hash TableLinked ListDesignDoubly-Linked ListHash MapDoubly Linked List

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n)	O(n)

💡

Intuition

Time O(1)Space O(n)

This approach uses a combination of a hash map and a doubly linked list to maintain the counts efficiently. The linked list allows for quick access to the minimum and maximum counts while the hash map keeps track of the keys and their corresponding nodes in the list.

⚙️

Algorithm

3 steps

1Step 1: Create a doubly linked list to maintain nodes representing counts.
2Step 2: Use a hash map to map each string to its count and the corresponding node in the linked list.
3Step 3: For increment and decrement operations, update the count and adjust the linked list accordingly to maintain order.

solution.py54 lines

1class Node:
2    def __init__(self, count):
3        self.count = count
4        self.keys = set()
5        self.prev = None
6        self.next = None
7
8class AllOne:
9    def __init__(self):
10        self.key_count = {}
11        self.head = Node(float('-inf'))
12        self.tail = Node(float('inf'))
13        self.head.next = self.tail
14        self.tail.prev = self.head
15
16    def _add_node(self, new_node, prev_node):
17        new_node.prev = prev_node
18        new_node.next = prev_node.next
19        prev_node.next.prev = new_node
20        prev_node.next = new_node
21
22    def _remove_node(self, node):
23        node.prev.next = node.next
24        node.next.prev = node.prev
25
26    def inc(self, key: str) -> None:
27        count = self.key_count.get(key, 0)
28        self.key_count[key] = count + 1
29        if count:
30            self._remove_node(self.key_count[key])
31        new_count = count + 1
32        if new_count not in self.key_count:
33            new_node = Node(new_count)
34            self._add_node(new_node, self.head)
35        self.head.next.keys.add(key)
36
37    def dec(self, key: str) -> None:
38        count = self.key_count[key]
39        self._remove_node(self.key_count[key])
40        if count == 1:
41            del self.key_count[key]
42        else:
43            self.key_count[key] = count - 1
44            new_count = count - 1
45            if new_count not in self.key_count:
46                new_node = Node(new_count)
47                self._add_node(new_node, self.head)
48            self.head.next.keys.add(key)
49
50    def getMaxKey(self) -> str:
51        return "" if self.head.next == self.tail else next(iter(self.head.next.keys))
52
53    def getMinKey(self) -> str:
54        return "" if self.tail.prev == self.head else next(iter(self.tail.prev.keys))

ℹ

Complexity note: The time complexity is O(1) for all operations because we use a hash map for key counts and a linked list for maintaining order, allowing for constant time access to max and min keys.

1Using a linked list allows for efficient insertions and deletions while maintaining order.
2A hash map provides O(1) access to the count of each key, which is crucial for performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.