leetcode - 380. Insert Delete GetRandom O(1)

Description

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.

Solution

I was thinking about using map to solve the problem, the time complexity for insert and delete could be $o(1)$, but for random function, since I have to transform the dict keys into list, it would take $o(n)$.

So in order to get $o(1)$ time for random, we better use list. For insert it’s easy to be $o(1)$, but for remove, the key point is that the only way to remove element from list in $o(1)$ is to pop. So when we need to remove an element, we could swap the element we need to remove with the last element in the list, and then we pop the last element from the list.

Code

class RandomizedSet:

    def __init__(self):
        self.vals = []
        self.map = {}
        self.index = 0

    def insert(self, val: int) -> bool:
        if val in self.map:
            return False
        self.map[val] = self.index
        self.vals.append(val)
        self.index += 1
        return True

    def remove(self, val: int) -> bool:
        if val not in self.map:
            return False
        # swap the last element with the one needs to be deleted
        self.vals[self.map[val]], self.vals[self.index - 1] = self.vals[self.index - 1], self.vals[self.map[val]]
        # use the index that was from the element to be deleted to update the last element, at this point, the last element is self.vals[self.map[val]] because of the swap
        self.map[self.vals[self.map[val]]] = self.map[val]
        self.index -= 1
        self.vals.pop()
        self.map.pop(val)
        return True

    def getRandom(self) -> int:
        return random.choice(self.vals)


# Your RandomizedSet object will be instantiated and called as such:
# obj = RandomizedSet()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()