Description
https://leetcode.com/problems/insert-delete-getrandom-o1/
Design a data structure that supports all following operations in average O(1) time.
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 (it’s guaranteed that at least one element exists when this method is called). 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();
Python Solution
Use hashmap and list together to support operations efficiently.
class RandomizedSet:
def __init__(self):
self.nums = []
self.val_to_index = {}
def insert(self, val: int) -> bool:
if val in self.val_to_index:
return False
self.nums.append(val)
self.val_to_index[val] = len(self.nums) - 1
return True
def remove(self, val: int) -> bool:
if val not in self.val_to_index:
return False
deleted_index = self.val_to_index[val]
if deleted_index < len(self.nums) - 1:
last_num = self.nums[-1]
self.nums[deleted_index] = last_num
self.val_to_index[last_num] = deleted_index
del self.val_to_index[val]
self.nums.pop()
return True
def getRandom(self) -> int:
return self.nums[random.randint(0, len(self.nums) - 1)]
# 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()
- Time Complexity: O(1)
- Space Complexity: O(N)
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