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OJ notes/pages/Leetcode Design-Hash.md
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OJ notes/pages/Leetcode Design-Hash.md
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# Leetcode Design-Hash
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#### 2022-09-01 14:44
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> ##### Data structures:
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> #DS #array #linked_list
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode #CS_list_need_practicing
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> ##### Revisions:
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> N/A
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##### Links:
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- [Link to problem](https://leetcode.com/problems/design-hashmap/)
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- [Explanation](https://leetcode.com/problems/design-hashmap/discuss/1097755/JS-Python-Java-C%2B%2B-or-(Updated)-Hash-and-Array-Solutions-w-Explanation)
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___
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### Problem
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Design a HashMap without using any built-in hash table libraries.
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Implement the `MyHashMap` class:
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- `MyHashMap()` initializes the object with an empty map.
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- `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`.
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- `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`.
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- `void remove(key)` removes the `key` and its corresponding `value` if the map contains the mapping for the `key`.
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#### Examples
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**Example 1:**
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**Input**
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["MyHashMap", "put", "put", "get", "get", "put", "get", "remove", "get"]
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[[], [1, 1], [2, 2], [1], [3], [2, 1], [2], [2], [2]]
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**Output**
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[null, null, null, 1, -1, null, 1, null, -1]
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**Explanation**
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MyHashMap myHashMap = new MyHashMap();
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myHashMap.put(1, 1); // The map is now [[1,1]]
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myHashMap.put(2, 2); // The map is now [[1,1], [2,2]]
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myHashMap.get(1); // return 1, The map is now [[1,1], [2,2]]
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myHashMap.get(3); // return -1 (i.e., not found), The map is now [[1,1], [2,2]]
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myHashMap.put(2, 1); // The map is now [[1,1], [2,1]] (i.e., update the existing value)
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myHashMap.get(2); // return 1, The map is now [[1,1], [2,1]]
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myHashMap.remove(2); // remove the mapping for 2, The map is now [[1,1]]
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myHashMap.get(2); // return -1 (i.e., not found), The map is now [[1,1]]
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#### Constraints
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- `0 <= key, value <= 106`
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- At most `104` calls will be made to `put`, `get`, and `remove`.
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### Thoughts
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> [!summary]
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> This is a #array problem.
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#### Easy way:
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Using array, since 10^6 size is not too large
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#### Hard way:
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Using hashes with linked list to support collision check.
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### Solution
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Easy way:
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```cpp
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class MyHashMap {
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int myMap[1000001];
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public:
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MyHashMap() { fill(myMap, myMap + 1000001, -1); }
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void put(int key, int value) { myMap[key] = value; }
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int get(int key) { return myMap[key]; }
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void remove(int key) { myMap[key] = -1; }
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};
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/**
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* Your MyHashMap object will be instantiated and called as such:
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* MyHashMap* obj = new MyHashMap();
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* obj->put(key,value);
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* int param_2 = obj->get(key);
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* obj->remove(key);
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*/
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```
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# {{title}}
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# {{title}}
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#### {{date}} {{time}}
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{{date}} {{time}}
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# {{title}}
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# {{title}}
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#### {{date}} {{time}}
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{{date}} {{time}}
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> ##### Algorithms:
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> ##### Algorithms:
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> #algorithm
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> #algorithm
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# {{title}}
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# {{title}}
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#### {{date}} {{time}}
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{{date}} {{time}}
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> ##### Algorithms:
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> ##### Algorithms:
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> #algorithm
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> #algorithm
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