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

In this guide, you will get 381. Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution with the best time and space complexity. The solution to Insert Delete GetRandom O() – Duplicates allowed problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Insert Delete GetRandom O() – Duplicates allowed solution in C++
Insert Delete GetRandom O() – Duplicates allowed solution in Java
Insert Delete GetRandom O() – Duplicates allowed solution in Python
Additional Resources

381. Insert Delete GetRandom O(1) - Duplicates allowed LeetCode Solution image

Problem Statement of Insert Delete GetRandom O() – Duplicates allowed

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 reporting 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 the 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.

Example 1:

Input
[“RandomizedCollection”, “insert”, “insert”, “insert”, “getRandom”, “remove”, “getRandom”]
[[], [1], [1], [2], [], [1], []]
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.

Constraints:

-231 <= val <= 231 – 1
At most 2 * 105 calls in total will be made to insert, remove, and getRandom.
There will be at least one element in the data structure when getRandom is called.

Complexity Analysis

Time Complexity: O(1)
Space Complexity: O(n)

381. Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution in C++

struct Item {
  int val;
  int indexInMap;
};

class RandomizedCollection {
 public:
  bool insert(int val) {
    valToIndices[val].push_back(items.size());
    items.emplace_back(val, valToIndices[val].size() - 1);
    return valToIndices[val].size() == 1;
  }

  bool remove(int val) {
    if (!valToIndices.contains(val))
      return false;

    const int index = valToIndices[val].back();
    valToIndices[items.back().val][items.back().indexInMap] = index;
    valToIndices[val].pop_back();
    if (valToIndices[val].empty())
      valToIndices.erase(val);
    items[index] = items.back();
    items.pop_back();
    return true;
  }

  int getRandom() {
    const int index = rand() % items.size();
    return items[index].val;
  }

 private:
  unordered_map<int, vector<int>> valToIndices;
  vector<Item> items;
};
/* code provided by PROGIEZ */

381. Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution in Java

class RandomizedCollection {
  public boolean insert(int val) {
    valToIndices.putIfAbsent(val, new ArrayList<>());
    valToIndices.get(val).add(items.size());
    items.add(new Item(val, valToIndices.get(val).size() - 1));
    return valToIndices.get(val).size() == 1;
  }

  public boolean remove(int val) {
    if (!valToIndices.containsKey(val))
      return false;

    final int index = lastIndex(valToIndices.get(val));
    valToIndices.get(last(items).val).set(last(items).indexInMap, index);
    final int indicesSize = valToIndices.get(val).size();
    valToIndices.get(val).remove(indicesSize - 1);
    if (valToIndices.get(val).isEmpty())
      valToIndices.remove(val);
    items.set(index, last(items));
    items.remove(items.size() - 1);
    return true;
  }

  public int getRandom() {
    final int index = rand.nextInt(items.size());
    return items.get(index).val;
  }

  private record Item(int val, int indexInMap) {}

  private Map<Integer, List<Integer>> valToIndices = new HashMap<>();
  private List<Item> items = new ArrayList<>();
  private Random rand = new Random();

  private int lastIndex(List<Integer> indices) {
    return indices.get(indices.size() - 1);
  }

  private Item last(List<Item> items) {
    return items.get(items.size() - 1);
  }
}
// code provided by PROGIEZ

381. Insert Delete GetRandom O(1) – Duplicates allowed LeetCode Solution in Python

class RandomizedCollection:
  def __init__(self):
    self.vals = []
    self.valToIndices = collections.defaultdict(list)

  def insert(self, val: int) -> bool:
    self.valToIndices[val].append(len(self.vals))
    self.vals.append([val, len(self.valToIndices[val]) - 1])
    return len(self.valToIndices[val]) == 1

  def remove(self, val: int) -> bool:
    if val not in self.valToIndices or self.valToIndices[val] == []:
      return False

    index = self.valToIndices[val][-1]
    self.valToIndices[self.vals[-1][0]][self.vals[-1][1]] = index
    self.valToIndices[val].pop()
    self.vals[index] = self.vals[-1]
    self.vals.pop()
    return True

  def getRandom(self) -> int:
    index = random.randint(0, len(self.vals) - 1)
    return self.vals[index][0]
 # code by PROGIEZ