All O`one Data Structure — LeetCode 432 Python Solution
- Problem
- #432
- Pattern
- Linked List
- Reading time
- 13 min
- Source
- leetcode.com
The problem
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.
Example
- Input
- ["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
- Output
- [null, null, null, "hello", "hello", null, "hello", "leet"]
- Explanation
- AllOne allOne = new AllOne();
Python solution
class Node:
def __init__(self, key='', cnt=0):
self.prev = None
self.next = None
self.cnt = cnt
self.keys = {key}
def insert(self, node):
node.prev = self
node.next = self.next
node.prev.next = node
node.next.prev = node
return node
def remove(self):
self.prev.next = self.next
self.next.prev = self.prev
class AllOne:
def __init__(self):
self.root = Node()
self.root.next = self.root
self.root.prev = self.root
self.nodes = {}
def inc(self, key: str) -> None:
root, nodes = self.root, self.nodes
if key not in nodes:
if root.next == root or root.next.cnt > 1:
nodes[key] = root.insert(Node(key, 1))
else:
root.next.keys.add(key)
nodes[key] = root.next
else:
curr = nodes[key]
next = curr.next
if next == root or next.cnt > curr.cnt + 1:
nodes[key] = curr.insert(Node(key, curr.cnt + 1))
else:
next.keys.add(key)
nodes[key] = next
curr.keys.discard(key)
if not curr.keys:
curr.remove()
def dec(self, key: str) -> None:
root, nodes = self.root, self.nodes
curr = nodes[key]
if curr.cnt == 1:
nodes.pop(key)
else:
prev = curr.prev
if prev == root or prev.cnt < curr.cnt - 1:
nodes[key] = prev.insert(Node(key, curr.cnt - 1))
else:
prev.keys.add(key)
nodes[key] = prev
curr.keys.discard(key)
if not curr.keys:
curr.remove()
def getMaxKey(self) -> str:
return next(iter(self.root.prev.keys))
def getMinKey(self) -> str:
return next(iter(self.root.next.keys))
# Your AllOne object will be instantiated and called as such:
# obj = AllOne()
# obj.inc(key)
# obj.dec(key)
# param_3 = obj.getMaxKey()
# param_4 = obj.getMinKey()Complexity
| Measure | Complexity |
|---|---|
| Time | O(n) |
| Space | O(n) auxiliary |
Pattern: Linked List
Rewire pointers in place, with a dummy head and a saved next to keep it safe. LeetCode 432. All O`one Data Structure is filed here because LeetCode tags it Linked List and Doubly-Linked List, which is the vocabulary this hub collects.
The linked list guide has the Python template for the pattern and the 75 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 432. All O`one Data Structure?
- LeetCode 432. All O`one Data Structure is rated Hard on LeetCode.
- What is the time complexity of LeetCode 432. All O`one Data Structure?
- The Python solution on this page runs in O(n).
- What is the space complexity of LeetCode 432. All O`one Data Structure?
- The Python solution on this page uses O(n) auxiliary space.
- What topics does LeetCode 432. All O`one Data Structure cover?
- LeetCode 432. All O`one Data Structure is tagged Design, Hash Table, Linked List and Doubly-Linked List on LeetCode.