432. All O`one Data Structure

Description

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.

Solutions

Solution 1

Python3Java

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()

class AllOne {
    Node root = new Node();
    Map<String, Node> nodes = new HashMap<>();

    public AllOne() {
        root.next = root;
        root.prev = root;
    }

    public void inc(String key) {
        if (!nodes.containsKey(key)) {
            if (root.next == root || root.next.cnt > 1) {
                nodes.put(key, root.insert(new Node(key, 1)));
            } else {
                root.next.keys.add(key);
                nodes.put(key, root.next);
            }
        } else {
            Node curr = nodes.get(key);
            Node next = curr.next;
            if (next == root || next.cnt > curr.cnt + 1) {
                nodes.put(key, curr.insert(new Node(key, curr.cnt + 1)));
            } else {
                next.keys.add(key);
                nodes.put(key, next);
            }
            curr.keys.remove(key);
            if (curr.keys.isEmpty()) {
                curr.remove();
            }
        }
    }

    public void dec(String key) {
        Node curr = nodes.get(key);
        if (curr.cnt == 1) {
            nodes.remove(key);
        } else {
            Node prev = curr.prev;
            if (prev == root || prev.cnt < curr.cnt - 1) {
                nodes.put(key, prev.insert(new Node(key, curr.cnt - 1)));
            } else {
                prev.keys.add(key);
                nodes.put(key, prev);
            }
        }

        curr.keys.remove(key);
        if (curr.keys.isEmpty()) {
            curr.remove();
        }
    }

    public String getMaxKey() {
        return root.prev.keys.iterator().next();
    }

    public String getMinKey() {
        return root.next.keys.iterator().next();
    }
}

class Node {
    Node prev;
    Node next;
    int cnt;
    Set<String> keys = new HashSet<>();

    public Node() {
        this("", 0);
    }

    public Node(String key, int cnt) {
        this.cnt = cnt;
        keys.add(key);
    }

    public Node insert(Node node) {
        node.prev = this;
        node.next = this.next;
        node.prev.next = node;
        node.next.prev = node;
        return node;
    }

    public void remove() {
        this.prev.next = this.next;
        this.next.prev = this.prev;
    }
}

/**
 * Your AllOne object will be instantiated and called as such:
 * AllOne obj = new AllOne();
 * obj.inc(key);
 * obj.dec(key);
 * String param_3 = obj.getMaxKey();
 * String param_4 = obj.getMinKey();
 */

432. All O`one Data Structure

Description

Solutions

Solution 1

Comments