Insert Delete GetRandom O(1)
Insert Delete GetRandom O(1):
Implement the RandomizedSet
class:
RandomizedSet()
Initializes theRandomizedSet
object.bool insert(int val)
Inserts an itemval
into the set if not present. Returnstrue
if the item was not present,false
otherwise.bool remove(int val)
Removes an itemval
from the set if present. Returnstrue
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 toinsert
,remove
, andgetRandom
. - 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:
Solution:
- Java
- JavaScript
- Python
- C++
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();
*/
var RandomizedSet = function () {
this.map = new Map();
this.list = [];
this.rnd = new Map();
};
/**
* @param {number} val
* @return {boolean}
*/
RandomizedSet.prototype.insert = function (val) {
if (this.map.has(val)) return false;
this.list.push(val);
this.map.set(val, this.list.length - 1);
return true;
};
/**
* @param {number} val
* @return {boolean}
*/
RandomizedSet.prototype.remove = function (val) {
if (!this.map.has(val)) return false;
const pos = this.map.get(val);
[this.list[pos], this.list[this.list.length - 1]] = [
this.list[this.list.length - 1],
this.list[pos],
];
this.map.set(this.list[pos], pos);
this.list.pop();
this.map.delete(val);
return true;
};
/**
* @return {number}
*/
RandomizedSet.prototype.getRandom = function () {
const randomIndex = Math.floor(Math.random() * this.list.length);
return this.list[randomIndex];
};
/**
* Your RandomizedSet object will be instantiated and called as such:
* var obj = new RandomizedSet()
* var param_1 = obj.insert(val)
* var param_2 = obj.remove(val)
* var param_3 = obj.getRandom()
*/
class RandomizedSet:
def __init__(self):
self.map = {}
self.list = []
self.rnd = random.Random()
def insert(self, val: int) -> bool:
if val in self.map:
return False
self.list.append(val)
self.map[val] = len(self.list) - 1
return True
def remove(self, val: int) -> bool:
if val not in self.map:
return False
pos = self.map[val]
self.list[pos], self.list[-1] = self.list[-1], self.list[pos]
self.map[self.list[pos]] = pos
self.list.pop()
del self.map[val]
return True
def getRandom(self) -> int:
randomIndex = self.rnd.randint(0, len(self.list) - 1)
return self.list[randomIndex]
# 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()
class RandomizedSet {
public:
RandomizedSet(): seed((random_device())()) {
}
bool insert(int val) {
if (map.count(val)) return false;
map[val] = list.size();
list.emplace_back(val);
return true;
}
bool remove(int val) {
if (!map.count(val)) return false;
list[map[val]] = list.back();
map[list.back()] = map[val];
list.pop_back();
map.erase(val);
return true;
}
int getRandom() {
return list[uniform_int_distribution<>
(0, list.size() - 1)(seed)];
}
private:
unordered_map<int, int> map;
vector<int> list;
default_random_engine seed;
};
/**
* Your RandomizedSet object will be instantiated and called as such:
* RandomizedSet* obj = new RandomizedSet();
* bool param_1 = obj->insert(val);
* bool 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 ourRandomizedSet
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.