Today I have done leetcode's April LeetCoding Challenge with cpp.

April LeetCoding Challenge 22

Description

Design HashMap

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 <= 10^6
At most 10^4 calls will be made to put, get, and remove.

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
constexpr int MOD = 4001;
class MyHashMap {
  vector<vector<pair<int, int>>> container;
public:
  /** Initialize your data structure here. */
  MyHashMap(): container(MOD) {}
  
  /** value will always be non-negative. */
  void put(int key, int value) {
    int val = key;
    key %= MOD;
    for(auto &[k, v] : container[key]) {
      if(k == val) {
        v = value;
        return;
      }
    }
    container[key].push_back({val, value});
  }
  
  /** Returns the value to which the specified key is mapped, or -1 if this map contains no mapping for the key */
  int get(int key) {
    int val = key;
    key %= MOD;
    for(auto [k, v] : container[key]) {
      if(k == val) return v;
    }
    return -1;
  }
  
  /** Removes the mapping of the specified value key if this map contains a mapping for the key */
  void remove(int key) {
    int val = key;
    key %= MOD;
    auto it = container[key].begin();
    while(it != container[key].end() && it->first != val) {
      ++it;
    }
    if(it != container[key].end()) {
      container[key].erase(it);
    }
  }
};

// Accepted
// 36/36 cases passed (110 ms)
// Your runtime beats 94.72 % of cpp submissions
// Your memory usage beats 60.03 % of cpp submissions (55.2 MB)

2022-04-22 Daily-Challenge

April LeetCoding Challenge 22

Description

Solution