LeetCode in Kotlin

460. LFU Cache

Hard

Design and implement a data structure for a Least Frequently Used (LFU) cache.

Implement the LFUCache class:

• LFUCache(int capacity) Initializes the object with the capacity of the data structure.
• int get(int key) Gets the value of the key if the key exists in the cache. Otherwise, returns -1.
• void put(int key, int value) Update the value of the key if present, or inserts the key if not already present. When the cache reaches its capacity, it should invalidate and remove the least frequently used key before inserting a new item. For this problem, when there is a tie (i.e., two or more keys with the same frequency), the least recently used key would be invalidated.

To determine the least frequently used key, a use counter is maintained for each key in the cache. The key with the smallest use counter is the least frequently used key.

When a key is first inserted into the cache, its use counter is set to 1 (due to the put operation). The use counter for a key in the cache is incremented either a get or put operation is called on it.

The functions get and put must each run in O(1) average time complexity.

Example 1:

Input

["LFUCache", "put", "put", "get", "put", "get", "get", "put", "get", "get", "get"] 
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [3], [4, 4], [1], [3], [4]]

Output: [null, null, null, 1, null, -1, 3, null, -1, 3, 4]

Explanation:

// cnt(x) = the use counter for key x 
   
// cache=[] will show the last used order for tiebreakers (leftmost element is most recent) 

LFUCache lfu = new LFUCache(2); 
lfu.put(1, 1); // cache=[1,_], cnt(1)=1
lfu.put(2, 2);  // cache=[2,1], cnt(2)=1, cnt(1)=1
lfu.get(1);     // return 1 
                // cache=[1,2], cnt(2)=1, cnt(1)=2
lfu.put(3, 3);  // 2 is the LFU key because cnt(2)=1 is the smallest, invalidate 2. 
                // cache=[3,1], cnt(3)=1, cnt(1)=2
lfu.get(2);     // return -1 (not found)
lfu.get(3);     // return 3
                // cache=[3,1], cnt(3)=2, cnt(1)=2
lfu.put(4, 4);  // Both 1 and 3 have the same cnt, but 1 is LRU, invalidate 1. 
                // cache=[4,3], cnt(4)=1, cnt(3)=2
lfu.get(1);     // return -1 (not found)
lfu.get(3);     // return 3 
                // cache=[3,4], cnt(4)=1, cnt(3)=3
lfu.get(4);     // return 4 
                // cache=[3,4], cnt(4)=2, cnt(3)=3

Constraints:

• 0 <= capacity <= 104
• 0 <= key <= 105
• 0 <= value <= 109
• At most 2 * 105 calls will be made to get and put.

Solution

class LFUCache(capacity: Int) {
    private class Node {
        var prev: Node? = null
        var next: Node? = null
        var key = -1
        var `val` = 0
        var freq = 0
    }

    private val endOfBlock: MutableMap<Int, Node?>
    private val map: MutableMap<Int, Node>
    private val capacity: Int
    private val linkedList: Node

    init {
        endOfBlock = HashMap()
        map = HashMap()
        this.capacity = capacity
        linkedList = Node()
    }

    operator fun get(key: Int): Int {
        if (map.containsKey(key)) {
            val newEndNode = map[key]
            val endNode: Node?
            val currEndNode = endOfBlock[newEndNode!!.freq]
            if (currEndNode === newEndNode) {
                findNewEndOfBlock(newEndNode)
                if (currEndNode.next == null || currEndNode.next!!.freq > newEndNode.freq + 1) {
                    newEndNode.freq++
                    endOfBlock[newEndNode.freq] = newEndNode
                    return newEndNode.`val`
                }
            }
            if (newEndNode.next != null) {
                newEndNode.next!!.prev = newEndNode.prev
            }
            newEndNode.prev!!.next = newEndNode.next
            newEndNode.freq++
            endNode = if (currEndNode!!.next == null || currEndNode.next!!.freq > newEndNode.freq) {
                currEndNode
            } else {
                endOfBlock[newEndNode.freq]
            }
            endOfBlock[newEndNode.freq] = newEndNode
            if (endNode!!.next != null) {
                endNode.next!!.prev = newEndNode
            }
            newEndNode.next = endNode.next
            endNode.next = newEndNode
            newEndNode.prev = endNode
            return newEndNode.`val`
        }
        return -1
    }

    fun put(key: Int, value: Int) {
        val endNode: Node?
        val newEndNode: Node
        if (capacity == 0) {
            return
        }
        if (map.containsKey(key)) {
            map[key]!!.`val` = value
            get(key)
        } else {
            if (map.size == capacity) {
                val toDelete = linkedList.next
                map.remove(toDelete!!.key)
                if (toDelete.next != null) {
                    toDelete.next!!.prev = linkedList
                }
                linkedList.next = toDelete.next
                if (endOfBlock[toDelete.freq] === toDelete) {
                    endOfBlock.remove(toDelete.freq)
                }
            }
            newEndNode = Node()
            newEndNode.key = key
            newEndNode.`val` = value
            newEndNode.freq = 1
            map[key] = newEndNode
            endNode = endOfBlock.getOrDefault(1, linkedList)
            endOfBlock[1] = newEndNode
            if (endNode!!.next != null) {
                endNode.next!!.prev = newEndNode
            }
            newEndNode.next = endNode.next
            endNode.next = newEndNode
            newEndNode.prev = endNode
        }
    }

    private fun findNewEndOfBlock(node: Node?) {
        val prev = node!!.prev
        if (prev!!.freq == node.freq) {
            endOfBlock[node.freq] = prev
        } else {
            endOfBlock.remove(node.freq)
        }
    }
}

/*
 * Your LFUCache object will be instantiated and called as such:
 * var obj = LFUCache(capacity)
 * var param_1 = obj.get(key)
 * obj.put(key,value)
 */

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