460. LFU Cache
Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following operations: get
and put
.
get(key)
- Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value)
- Set or insert the value if the key is not already present. When the cache reaches its capacity, it should invalidate the least frequently used item before inserting a new item. For the purpose of this problem, when there is a tie (i.e., two or more keys that have the same frequency), the least recently used key would be evicted.
Follow up: Could you do both operations in O(1) time complexity?
Example:
LFUCache cache = new LFUCache( 2 /* capacity */ );
cache.put(1, 1);
cache.put(2, 2);
cache.get(1); // returns 1
cache.put(3, 3); // evicts key 2
cache.get(2); // returns -1 (not found)
cache.get(3); // returns 3.
cache.put(4, 4); // evicts key 1.
cache.get(1); // returns -1 (not found)
cache.get(3); // returns 3
cache.get(4); // returns 4
class LFUCache {
public:
LFUCache(int capacity) : cap(capacity), minFreq(0) {
}
int get(int key) {
if (!m.count(key)) return -1;
freq[m[key].second].erase(iter[key]);
++m[key].second; // increment the frequency of the key
freq[m[key].second].push_back(key);
iter[key] = --freq[m[key].second].end();
if (freq[minFreq].empty()) ++minFreq;
return m[key].first;
}
void put(int key, int value) {
if (cap <= 0) return;
if (get(key) != -1) {
m[key].first = value;
return;
}
if (m.size() >= cap) {
m.erase(freq[minFreq].front());
iter.erase(freq[minFreq].front());
freq[minFreq].pop_front();
}
m[key] = {value, 1};
freq[1].push_back(key);
iter[key] = --freq[1].end();
minFreq = 1;
}
private:
int cap;
int minFreq;
unordered_map<int, pair<int, int> > m; // key -> <value, freq>
unordered_map<int, list<int> > freq; // freq -> list of keys with the same frequency
unordered_map<int, list<int>::iterator> iter; // key -> iterator pointing to the place in the freq list
};
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