706

706. Design HashMap (Easy)

Design a HashMap without using any built-in hash table libraries.

Implement the MyHashMap class:

• MyHashMap() initializes the object with an empty map.
• void put(int key, int value) inserts a (key, value) pair into the HashMap. If the key already exists in the map, update the corresponding value.
• int get(int key) returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key.
• void remove(key) removes the key and its corresponding value if the map contains the mapping for the key.

 

Example 1:

Input
["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
Output
[null, null, null, 1, -1, null, 1, null, -1]

Explanation
MyHashMap myHashMap = new MyHashMap();
myHashMap.put(1, 1); // The map is now [[1,1]]
myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
myHashMap.get(1);    // return 1, The map is now [[1,1], [2,2]]
myHashMap.get(3);    // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
myHashMap.get(2);    // return 1, The map is now [[1,1], [2,1]]
myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
myHashMap.get(2);    // return -1 (i.e., not found), The map is now [[1,1]]

 

Constraints:

• 0 <= key, value <= 106
• At most 104 calls will be made to put, get, and remove.

Companies:
Amazon, Microsoft, LinkedIn, Salesforce, Oracle, Apple, ServiceNow, Goldman Sachs, Walmart Global Tech, Bloomberg

Related Topics:
Array, Hash Table, Linked List, Design, Hash Function

Similar Questions:

• Design HashSet (Easy)
• Design Skiplist (Hard)

Solution 1.

// OJ: https://leetcode.com/problems/design-hashmap/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class MyHashMap {
    const static int N = 1000001;
    int m[N];
public:
    MyHashMap() {
        for (int i = 0; i < N; ++i) m[i] = -1;
    }
    void put(int key, int value) {
        m[key] = value;
    }
    int get(int key) {
        return m[key];
    }
    void remove(int key) {
        m[key] = -1;
    }
};

Solution 2. Modulo + Doubly Linked List

// OJ: https://leetcode.com/problems/design-hashmap/
// Author: github.com/lzl124631x
// Time:
//		MyHashMap: O(1)
//		put, get, remove: O(logN) on average, O(N) in the worst case.
// Space: O(N)
const int keyRange = 769;
class Bucket {
    list<pair<int, int>> items;
    list<pair<int, int>>::iterator find(int key) {
        auto it = begin(items);
        while (it != end(items) && it->first != key) ++it;
        return it;
    }
public:
    Bucket() {}
    void put(int key, int val) {
        auto it = find(key);
        if (it == end(items)) items.emplace_front(key, val);
        else it->second = val;
    }
    int get(int key) {
        auto it = find(key);
        return it != end(items) ? it->second : -1;
    }
    void remove(int key) {
        auto it = find(key);
        if (it != end(items)) items.erase(it);
    }
};
class MyHashMap {
    Bucket buckets[keyRange] = {};
    int hash(int key) { return key % keyRange; };
public:
    MyHashMap() {
    }
    void put(int key, int value) {
        buckets[hash(key)].put(key, value);
    }
    int get(int key) {
        return buckets[hash(key)].get(key);
    }
    void remove(int key) {
        buckets[hash(key)].remove(key);
    }
};

Solution 3. Modulo + Binary Search Tree

// OJ: https://leetcode.com/problems/design-hashmap
// Author: github.com/lzl124631x
// Time:
//		MyHashMap: O(1)
//		put, get, remove: O(logN) on average, O(N) in the worst case.
// Space: O(N)
const int keyRange = 769;
struct BSTreeNode {
    int key, val;
    BSTreeNode *left = nullptr, *right = nullptr;
    BSTreeNode(int key, int val) : key(key), val(val) {}
};
class BSTree {
    BSTreeNode *root = nullptr;
    BSTreeNode* deleteNode(BSTreeNode* root, int key) {
        if (!root) return nullptr;
        if (root->key < key) {
            root->right = deleteNode(root->right, key);
            return root;
        } else if (root->key > key) {
            root->left = deleteNode(root->left, key);
            return root;
        }
        if (!root->left || !root->right) return root->left ? root->left : root->right;
        auto newRoot = root->right, left = newRoot->left, node = root->left;
        newRoot->left = root->left;
        while (node->right) node = node->right;
        node->right = left;
        return newRoot;
    }
public:
    BSTree() {}
    void put(int key, int val) {
       if (!root) {
            root = new BSTreeNode(key, val);
            return;
        }
        BSTreeNode *prev = nullptr, *n = root;
        while (n) {
            prev = n;
            if (key == n->key) {
                n->val = val;
                return;
            } else if (key < n->key) n = n->left;
            else n = n->right;
        }
        if (key < prev->key) prev->left = new BSTreeNode(key, val);
        else prev->right = new BSTreeNode(key, val);
    }
    int get(int key) {
        auto n = root;
        while (n) {
            if (key == n->key) return n->val;
            if (key < n->key) n = n->left;
            else n = n->right;
        }
        return -1;
    }
    void remove(int key) {
        root = deleteNode(root, key);
    }
};
class MyHashMap {
    BSTree buckets[keyRange] = {};
    int hash(int key) { return key % keyRange; };
public:
    MyHashMap() {
    }
    void put(int key, int value) {
        buckets[hash(key)].put(key, value);
    }
    int get(int key) {
        return buckets[hash(key)].get(key);
    }
    void remove(int key) {
        buckets[hash(key)].remove(key);
    }
};