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Insert Delete GetRandom O(1)


Insert Delete GetRandom O(1): Implement the RandomizedSet class:

  • RandomizedSet() Initializes the RandomizedSet object.
  • bool insert(int val) Inserts an item val into the set if not present. Returns true if the item was not present, false otherwise.
  • bool remove(int val) Removes an item val from the set if present. Returns true if the item was present, false otherwise.
  • int getRandom() Returns a random element from the current set of elements (it's guaranteed that at least one element exists when this method is called). Each element must have the same probability of being returned.

You must implement the functions of the class such that each function works in average O(1) time complexity.

Example 1:
Input
["RandomizedSet", "insert", "remove", "insert", "getRandom", "remove",
"insert", "getRandom"]
[[], [1], [2], [2], [], [1], [2], []]
Output
[null, true, false, true, 2, true, false, 2]

Explanation
RandomizedSet randomizedSet = new RandomizedSet();
randomizedSet.insert(1); // Inserts 1 to the set. Returns true as 1
was inserted successfully.
randomizedSet.remove(2); // Returns false as 2 does not exist in the set.
randomizedSet.insert(2); // Inserts 2 to the set, returns true. Set now contains [1,2].
randomizedSet.getRandom(); // getRandom() should return either 1 or 2 randomly.
randomizedSet.remove(1); // Removes 1 from the set, returns true. Set now contains [2].
randomizedSet.insert(2); // 2 was already in the set, so return false.
randomizedSet.getRandom(); // Since 2 is the only number in the set,
getRandom() will always return 2.

Constraints:
  • -2^31 <= val <= 2^31 - 1
  • At most 2 * 10^5 calls will be made to insert, remove, and getRandom.
  • There will be at least one element in the data structure when getRandom is called.

Try this Problem on your own or check similar problems:

  1. Insert Delete GetRandom O(1) - Duplicates allowed
Solution:
class RandomizedSet {
Map<Integer, Integer> map;
List<Integer> list;
Random rnd;
public RandomizedSet() {
rnd = new Random();
list = new ArrayList<>();
map = new HashMap<>();
}

public boolean insert(int val) {
if(map.containsKey(val)) return false;
list.add(val);
map.put(val, list.size() - 1);
return true;
}

public boolean remove(int val) {
if(!map.containsKey(val)) return false;
Integer pos = map.get(val);
Collections.swap(list, pos, list.size() - 1);
map.put(list.get(pos), pos);
list.remove(list.size() - 1);
map.remove(val);
return true;
}

public int getRandom() {
return list.get(rnd.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/Space Complexity:
  • Time Complexity: O(1)
  • Space Complexity: O(n) where n is the number of elements added to our RandomizedSet

Explanation:

We have constant time complexity with linear space complexity proportional to the number of elements added to the RandomizedSet. We maintain a list and keep adding the numbers to it, while tracking the index of the value with map, so we don't have to search it linearly when asked to remove. getRandom just takes a random index from 0 inclusive to n (exclusive). The main trick is in the remove function, since removal from random position in list will cost us O(n) (we must shift elements), we can just always remove from the end. We first get the position of the val using our map, then we swap val with the last element. Now we have a new value at the val position so we have to update map map.put(list.get(pos), pos);, finally we remove val both from the map and from the end of the list. This approach leads to an average O(1) time complexity.