Problem 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.

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.

https://leetcode.com/problems/insert-delete-getrandom-o1/

Examples

Exampple 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.

How to Solve

The problem asks O(1) time complexity, so a hash table or set is the data structure to use here. Both insert and remove can be done by the hash table (or set) only. However, the getRandom method needs index access. The index access is not easy to implement by the hash table (or set) for most languages which don’t keep keys in ordered manner. To make the index access possible, the solution here adds an array as well. The hash table maintains a value to array’s index map. The array maintains values so far. The insert method adds value to both hash table and array. The remove method replaces the given value to array’s last element. This way, we can implement methods by O(1) time complexity.

Solution

• C++
• Java
• JavaScript
• Python
• Ruby

• class RandomizedSet {
  private:
      unordered_map<int, int> valToIdx;
      vector<int> values;
  
  public:
      RandomizedSet() {
          ios::sync_with_stdio(false);
          cin.tie(0);
          cout.tie(0);
      }
  
      bool insert(int val) {
          if (valToIdx.find(val) != valToIdx.end()) {
              return false;
          }
          valToIdx.insert({val, values.size()});
          values.push_back(val);
          return true;
      }
  
      bool remove(int val) {
          if (valToIdx.find(val) == valToIdx.end()) {
              return false;
          }
          int idx = valToIdx[val];
          int last = values.back();
          valToIdx[last] = idx;
          values[idx] = last;
          valToIdx.erase(val);
          values.pop_back();
          return true;
      }
  
      int getRandom() {
          return values[rand() % values.size()];
      }
  };
  

• 
  var RandomizedSet = function() {
      this.valToIdx = new Map();
      this.values = [];
  };
  
  /** 
   * @param {number} val
   * @return {boolean}
   */
  RandomizedSet.prototype.insert = function(val) {
      if (this.valToIdx.has(val)) {
          return false;
      }
      this.valToIdx.set(val, this.values.length);
      this.values.push(val);
      return true;
  };
  
  /** 
   * @param {number} val
   * @return {boolean}
   */
  RandomizedSet.prototype.remove = function(val) {
      if (!this.valToIdx.has(val)) {
          return false;
      }
      let idx = this.valToIdx.get(val);
      let last = this.values.at(-1);
      this.valToIdx.set(last, idx)
      this.values[idx] = last;
      this.valToIdx.delete(val);
      this.values.pop();
      return true;
  }
  
  /**
   * @return {number}
   */
  RandomizedSet.prototype.getRandom = function() {
      return this.values[Math.floor(Math.random() * this.values.length)];
  };
  

• import random
  
  class RandomizedSet:
  
      def __init__(self):
          self.valToIdx, self.values = {}, []
  
  
      def insert(self, val: int) -> bool:
          if val in self.valToIdx:
              return False
          self.valToIdx[val] = len(self.values)
          self.values.append(val)
          return True
  
  
      def remove(self, val: int) -> bool:
          if val not in self.valToIdx:
              return False
          idx, last = self.valToIdx[val], self.values[-1]
          self.valToIdx[last] = idx
          self.values[idx] = last
          del self.valToIdx[val]
          self.values.pop();
          return True
  
  
      def getRandom(self) -> int:
          return random.choice(self.values)
  

Complexities

• Time: O(1)
• Space: O(n)