# Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution

Difficulty Level Hard

## Problem Statement:

Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution: `RandomizedCollection` is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also removing a random element.

Implement the `RandomizedCollection` class:

• `RandomizedCollection()` Initializes the empty `RandomizedCollection` object.
• `bool insert(int val)` Inserts an item `val` into the multiset, even if the item is already present. Returns `true` if the item is not present, `false` otherwise.
• `bool remove(int val)` Removes an item `val` from the multiset if present. Returns `true` if the item is present, `false` otherwise. Note that if `val` has multiple occurrences in the multiset, we only remove one of them.
• `int getRandom()` Returns a random element from the current multiset of elements. The probability of each element being returned is linearly related to the number of same values the multiset contains.

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

Note: The test cases are generated such that `getRandom` will only be called if there is at least one item in the `RandomizedCollection`.

## Examples:

Input

```["RandomizedCollection", "insert", "insert", "insert", "getRandom", "remove", "getRandom"]
[[], , , , [], , []]
```

Output

```[null, true, false, true, 2, true, 1]

```

Explanation

```RandomizedCollection randomizedCollection = new RandomizedCollection();
randomizedCollection.insert(1);   // return true since the collection does not contain 1.
// Inserts 1 into the collection.
randomizedCollection.insert(1);   // return false since the collection contains 1.
// Inserts another 1 into the collection. Collection now contains [1,1].
randomizedCollection.insert(2);   // return true since the collection does not contain 2.
// Inserts 2 into the collection. Collection now contains [1,1,2].
randomizedCollection.getRandom(); // getRandom should:
// - return 1 with probability 2/3, or
// - return 2 with probability 1/3.
randomizedCollection.remove(1);   // return true since the collection contains 1.
// Removes 1 from the collection. Collection now contains [1,2].
randomizedCollection.getRandom(); // getRandom should return 1 or 2, both equally likely.```

## Approach:

### Idea:

The data structure can be implemented with the help of a `Hash_map` consisting of a pair of `<int, Hash_set>` and an `arrayList`. The `Hash_map` will store the indices of the elements and the `list` will store the list of elements present so far. The required methods can be implemented as follows:

• `insert`: Insert the element to the `list` and add the index to `HashMap[element]`.
• `remove`: First, we find any index of the element to be removed in the `HashMap`. Now we will update the element in this index with the last element of the `list`. Since the last element of the `list` has been assigned a new index, we have to update its value in the `HashMap`. We also have to get rid of the index of the element we removed from the `HashMap`.
• `getRandom`: Just sample a random element from the list.

### Code for Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution:

C++ Solution:

```class RandomizedCollection {
public:
unordered_map<int,unordered_set<int>> mp;
vector<int> arr;
int size = 0;

RandomizedCollection() {

}

bool insert(int val) {
bool status = not mp.count(val);
mp[val].insert(size);
arr.push_back(val);
size++;
return status;
}

bool remove(int val) {
bool status = mp.count(val);
if(status){
int last_element = arr.back();
if(val == last_element){
mp[val].erase(arr.size()-1);
arr.pop_back();
size--;
if(mp[val].size()==0)
mp.erase(val);
}
else{
int index;
for(auto it:mp[val]){
index = it;
break;
}
arr[index] = last_element;
mp[last_element].erase(arr.size()-1);
arr.pop_back();
size--;
mp[val].erase(index);
mp[last_element].insert(index);
if(mp[val].size()==0)
mp.erase(val);
}
}
return status;
}

int getRandom() {
return arr[rand()%arr.size()];
}
};

/**
* Your RandomizedCollection object will be instantiated and called as such:
* RandomizedCollection* obj = new RandomizedCollection();
* bool param_1 = obj->insert(val);
* bool param_2 = obj->remove(val);
* int param_3 = obj->getRandom();
*/```

## Complexity Analysis of Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution:

• Time Complexity: The time complexity of the above code is O(n) with n being the number of operations.
• Space ComplexityThe space complexity of the above code is O(n).
Translate »