## Sunday, August 7, 2016

### LeetCode 380 - Insert Delete GetRandom O(1)

http://www.cnblogs.com/grandyang/p/5740864.html
Design a data structure that supports all following operations in average O(1) time.

1. `insert(val)`: Inserts an item val to the set if not already present.
2. `remove(val)`: Removes an item val from the set if present.
3. `getRandom`: Returns a random element from current set of elements. Each element must have the same probability of being returned.

Example:
```// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();

// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);

// Returns false as 2 does not exist in the set.
randomSet.remove(2);

// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);

// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();

// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);

// 2 was already in the set, so return false.
randomSet.insert(2);

// Since 1 is the only number in the set, getRandom always return 1.
randomSet.getRandom();```

ArrayList 的本质是 Array,
so list.remove(list.size() - 1) -> time complexity is only O(1)
``````public class RandomizedSet {
private HashMap<Integer, Integer> map; // value -> index
private ArrayList<Integer> list; // value
private Random rand;
/** Initialize your data structure here. */
public RandomizedSet() {
map = new HashMap<Integer, Integer>();
list = new ArrayList<Integer>();
rand = new Random();
}

/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
if (map.containsKey(val)) {
return false;
}
else {
map.put(val, list.size());
return true;
}
}

/** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
if (!map.containsKey(val)) {
return false;
}
else {
int index = map.remove(val);
int last = list.remove(list.size() - 1);
if (last != val) {
list.set(index, last);
map.put(last, index);
}
return true;
}
}

/** Get a random element from the set. */
public int getRandom() {
int index = rand.nextInt(list.size());
return list.get(index);
}
}``````
X.
``````    private HashMap<Integer, Integer> keyMap = null;
private HashMap<Integer, Integer> valueMap = null;
int count;

/** Initialize your data structure here. */
public RandomizedSet() {
keyMap = new HashMap<Integer, Integer>();
valueMap = new HashMap<Integer, Integer>();
}

/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
if(keyMap.containsKey(val)) {
return false;
} else {
keyMap.put(val, count);
valueMap.put(count, val);
count = keyMap.size();
return true;
}
}

/** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
if(!keyMap.containsKey(val)) {
return false;
} else {
int valueKey = keyMap.get(val);
keyMap.remove(val);
if(valueKey != valueMap.size() - 1) {
valueMap.put(valueKey, valueMap.get(valueMap.size() - 1));
keyMap.put(valueMap.get(valueMap.size() - 1), valueKey);
valueMap.remove(valueMap.size() - 1);
} else {
valueMap.remove(valueKey);
}
count = keyMap.size();
return true;
}
}

/** Get a random element from the set. */
public int getRandom() {
Random random = new Random();
int n = random.nextInt(keyMap.size());
return valueMap.get(n);
}``````
We can use two hashmaps to solve this problem. One uses value as keys and the other uses index as the keys.
``` public class RandomizedSet {   HashMap<Integer, Integer> map1; HashMap<Integer, Integer> map2; Random rand;   /** Initialize your data structure here. */ public RandomizedSet() { map1 = new HashMap<Integer, Integer>(); map2 = new HashMap<Integer, Integer>(); rand = new Random(System.currentTimeMillis()); }   /** Inserts a value to the set. Returns true if the set did not already contain the specified element. */ public boolean insert(int val) { if(map1.containsKey(val)){ return false; }else{ map1.put(val, map1.size()); map2.put(map2.size(), val); } return true; }   /** Removes a value from the set. Returns true if the set contained the specified element. */ public boolean remove(int val) { if(map1.containsKey(val)){ int index = map1.get(val);   //remove the entry from both maps map1.remove(val); map2.remove(index);   if(map1.size()==0){ return true; }   //if last is deleted, do nothing if(index==map1.size()){ return true; }   //update the last element's index int key1 = map2.get(map2.size());   map1.put(key1, index); map2.remove(map2.size()); map2.put(index, key1);   }else{ return false; }   return true; }   /** Get a random element from the set. */ public int getRandom() { if(map1.size()==0){ return -1; }   if(map1.size()==1){ return map2.get(0); }   return map2.get(new Random().nextInt(map1.size())); //return 0; } } ```
https://discuss.leetcode.com/topic/53216/java-solution-using-a-hashmap-and-an-arraylist-along-with-a-follow-up-131-ms/4
How do you modify your code to allow duplicated number?
The follow-up: allowing duplications.
For example, after insert(1), insert(1), insert(2), getRandom() should have 2/3 chance return 1 and 1/3 chance return 2.
Then, remove(1), 1 and 2 should have an equal chance of being selected by getRandom().
The idea is to add a set to the hashMap to remember all the locations of a duplicated number.
``````public class RandomizedSet {
ArrayList<Integer> nums;
HashMap<Integer, Set<Integer>> locs;
java.util.Random rand = new java.util.Random();
/** Initialize your data structure here. */
public RandomizedSet() {
nums = new ArrayList<Integer>();
locs = new HashMap<Integer, Set<Integer>>();
}

/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
public boolean insert(int val) {
boolean contain = locs.containsKey(val);
if ( ! contain ) locs.put( val, new HashSet<Integer>() );
return ! contain ;
}

/** Removes a value from the set. Returns true if the set contained the specified element. */
public boolean remove(int val) {
boolean contain = locs.containsKey(val);
if ( ! contain ) return false;
int loc = locs.get(val).iterator().next();
if (loc < nums.size() - 1 ) {
int lastone = nums.get(nums.size() - 1 );
nums.set( loc , lastone );
locs.get(lastone).remove(nums.size() - 1);
}
nums.remove(nums.size() - 1);
locs.get(val).remove(loc);
if (locs.get(val).isEmpty()) locs.remove(val);
return true;
}

/** Get a random element from the set. */
public int getRandom() {
return nums.get( rand.nextInt(nums.size()) );
}
}``````
2 Hash Table + 1 List
valueMap<Integer, Integer> stores key-value pair
indexMap<Integer, Integer> stores key-index pair, where index is the index in the List. This hash table is to support delete operation in O(1) time.
List is for getRandom() method.
There is another tricky place in Delete operation. Instead of compare the trade off between ArrayList and LinkedList, maintain a variable N to store the number of keys in this data structure, whenever a delete operation is executed, just swap the keys with the last one and decrement N by one. In this way, delete operation is guaranteed in Constant time.
`class` `RandomizedHashTable {`
`  `
`  ``Map<Integer, Integer> valueMap = ``new` `HashMap<>();`
`  ``Map<Integer, Integer> indexMap = ``new` `HashMap<>();`
`  ``List<Integer> keys = ``new` `ArrayList<>();`
`  ``int` `N = ``0``;`
`  `
`  ``/* Insert a key-value */`
`  ``public` `void` `insert(``int` `key, ``int` `val) {`
`    ``if` `(valueMap.containsKey(key)) {`
`      ``return``;`
`    ``}`
`    `
`    ``valueMap.put(key, val);`
`    ``keys.add(key);`
`    ``indexMap.put(key, keys.size() - ``1``);`
`    ``N++;`
`  ``}`
`  `
`  ``/* Get the value */`
`  ``public` `int` `get(``int` `key) {`
`    ``if` `(!valueMap.containsKey(key)) {`
`      ``throw` `new` `IllegalArgumentException();`
`    ``}`
`    `
`    ``return` `valueMap.get(key);`
`  ``}`
`  `
`  ``/* Update the value */`
`  ``public` `void` `set(``int` `key, ``int` `val) {`
`    ``if` `(!valueMap.containsKey(key)) {`
`      ``return``;`
`    ``}`
`    `
`    ``valueMap.put(key, val);`
`  ``}`
`  `
`  ``/* Get a random value */`
`  ``public` `int` `getRandom() {`
`    ``int` `index = ``new` `Random().randInt(keys.size());`
`    ``return` `valueMap.get(keys.get(index));`
`  ``}`
`  `
`  ``/* Delete a key-value */`
`  ``public` `void` `delete(``int` `key) {`
`    ``if``(!valueMap.containsKey(key)) {`
`      ``return``;`
`    ``}`
`    `
`    ``valueMap.remove(key); // ALSO `indexMap.remove(key);
`    ``int` `index = indexMap.get(key);`
`    ``int` `last = keys.get(N - ``1``);`
`    ``indexMap.put(last, index);`
`    ``swap(keys, index, N - ``1``);`
`    ``N--;`
`  ``}`
`  `
`  ``private` `void` `swap(List<Integer> keys, ``int` `l, ``int` `r) {`
`    ``int` `tmp = keys.get(l);`
`    ``keys.set(l, keys.get(r);`
`    ``keys.set(r, tmp);`
`  ``}`
`}`

Design a data structure to support insert(), delete(), medium() and mode()
mode() 是众数，出现最多的那个。
BST + Heap
class TNode {
int val;
int freq;
int leftSize;
}
insert() O(logn)
delete() O(logn)
medium() O(logn)
mode() O(1)
Use a PriorityQueue to on TNode.freq to get mode number.
Implement a load balancer for web servers. It provide the following functionality:
Remove a bad server from the cluster => remove(server_id).
Pick a server in the cluster randomly with equal probability => pick().
Example
At beginning, the cluster is empty => {}.
pick()
>> 1         // the return value is random, it can be either 1, 2, or 3.
pick()
>> 2
pick()
>> 1
pick()
>> 3
remove(1)
pick()
>> 2
pick()
>> 3
pick()
>> 3

1，如何O(1)的根据server_id删除某一个server。
2，如何保证pick概率平均。

1，为了保证O(1)删除，那么首先一定不能使用ArrayList或者数组之类的结构，因为查找server_id必须要遍历。就算根据server_id进行了排序，最快也得O(log(n))。
可以做到O(1)删除的数据结构最容易想到的是HashMap和HashSet。
2，为了保证pick概率平均，最简单的方案是round robin。用一个pointer轮流指向并返回列表中的server。到达最后一个的时后就返回列表头。

*/

private class Node {
int id;
Node prev, next;
Node(int id) {
this.id = id;
prev = this;
next = this;
}
}

Map<Integer, Node> servers;
Node server;

servers = new HashMap<>();
server = null;
}

// @param server_id add a new server to the cluster
// @return void
if (servers.containsKey(server_id)) {
return;
}

Node newServer = new Node(server_id);
if (server == null) {
// initiate first server
server = newServer;
}

newServer.next = server.next;
newServer.prev = server;
server.next.prev = newServer;
server.next = newServer;

servers.put(server_id, newServer);
}

// @param server_id server_id remove a bad server from the cluster
// @return void
public void remove(int server_id) {
if (!servers.containsKey(server_id)) {
return;
}

Node node = servers.get(server_id);
if (node == server) {
// current server will be removed. move to next server.
server = server.next;
}

node.prev.next = node.next;
node.next.prev = node.prev;

servers.remove(server_id);
if (servers.isEmpty()) {
// if last server is removed, clear pointer
server = null;
}
}

// @return pick a server in the cluster randomly with equal probability
public int pick() {
server = server.next;
return server.id;
}