[LeetCode] 380. Insert Delete GeRandom O(1)
Design a data structure that supports all following operations in average O(1) time.
Thought process:
Because the size of the randomized set is always changing, getRandom requires a data structure that can map index to value. I will use a hash table to check if it contains certain value in constant time, and use an array to return random value in constant time. The hash map maps integer values to their positions in the array.
Solution:
Time complexity:
insert(val)
: Inserts an item val to the set if not already present.remove(val)
: Removes an item val from the set if present.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 2 is the only number in the set, getRandom always return 2. randomSet.getRandom();
Because the size of the randomized set is always changing, getRandom requires a data structure that can map index to value. I will use a hash table to check if it contains certain value in constant time, and use an array to return random value in constant time. The hash map maps integer values to their positions in the array.
- insert: check if the value already exists. If so return false. Else, add the value to list. Put value -> index mapping into map.
- remove: check if value exists. If not, return false. Else, get its index in the list. Swap it with the last element in the list. Update the mappings. Remove the last element from the list and map.
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 | class RandomizedSet { private List<Integer> list; private Map<Integer, Integer> map; /** Initialize your data structure here. */ public RandomizedSet() { list = new ArrayList<>(); map = new HashMap<>(); } /** 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; } list.add(val); map.put(val, list.size() - 1); 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; } int index = map.get(val); int last = list.size() - 1; list.set(index, list.get(last)); map.put(list.get(index), index); list.remove(last); map.remove(val); return true; } /** Get a random element from the set. */ public int getRandom() { Random random = new Random(); return list.get(random.nextInt(list.size())); } } /** * Your RandomizedSet object will be instantiated and called as such: * RandomizedSet obj = new RandomizedSet(); * boolean param_1 = obj.insert(val); * boolean param_2 = obj.remove(val); * int param_3 = obj.getRandom(); */ |
Time complexity:
- insert: O(1).
- remove: O(1).
- getRandom: O(1).
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