2023-10-04 Daily Challenge
Today I have done leetcode's October LeetCoding Challenge with cpp.
October LeetCoding Challenge 4
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 thekeyalready exists in the map, update the correspondingvalue.int get(int key)returns thevalueto which the specifiedkeyis mapped, or-1if this map contains no mapping for thekey.void remove(key)removes thekeyand its correspondingvalueif the map contains the mapping for thekey.
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
104calls will be made toput,get, andremove.
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 (95 ms)
// Your runtime beats 92.1 % of cpp submissions
// Your memory usage beats 59.66 % of cpp submissions (55.2 MB)