solution added.

Author: 149ps

30 Days Challange/Week 4/3. LRU Cache/solution.py

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+"""
+Design and implement a data structure for Least Recently Used (LRU) 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 reached its capacity, it should invalidate the least recently used item before inserting a new item.
+
+The cache is initialized with a positive capacity.
+
+Follow up:
+Could you do both operations in O(1) time complexity?
+
+Example:
+
+LRUCache cache = new LRUCache( 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.put(4, 4);    // evicts key 1
+cache.get(1);       // returns -1 (not found)
+cache.get(3);       // returns 3
+cache.get(4);       // returns 4
+"""
+
+
+class LRUCache:
+
+    def __init__(self, capacity: int):
+        self.hmap = collections.OrderedDict()
+        self.capacity = capacity
+
+    def get(self, key: int) -> int:
+        if key not in self.hmap.keys():
+            return -1
+        value = self.hmap.pop(key)
+        self.hmap[key] = value
+        return value
+
+    def put(self, key: int, value: int) -> None:
+        if key in self.hmap.keys():
+            self.hmap.pop(key)
+        else:
+            if len(self.hmap) >= self.capacity:
+                self.hmap.popitem(last=False)
+        self.hmap[key] = value
+
+# Your LRUCache object will be instantiated and called as such:
+# obj = LRUCache(capacity)
+# param_1 = obj.get(key)
+# obj.put(key,value)