How do you solve Insert Delete GetRandom O(1) - Duplicates allowed on LeetCode?

Insert Delete GetRandom O(1) - Duplicates allowed (LeetCode #381) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) on average for all operations time complexity and O(n) space complexity. Key insight: Using a hash map allows for efficient index tracking.

What is the time complexity of Insert Delete GetRandom O(1) - Duplicates allowed?

The optimal solution for Insert Delete GetRandom O(1) - Duplicates allowed runs in O(1) on average for all operations time and uses O(n) space. The use of a list allows O(1) access for random elements, while the hash map allows O(1) average time complexity for insertions and removals by keeping track of indices.

What is the brute force solution for Insert Delete GetRandom O(1) - Duplicates allowed?

The brute force approach for Insert Delete GetRandom O(1) - Duplicates allowed is Brute Force, with O(n²) time complexity and O(1) space complexity. In a brute-force approach, we can maintain a simple list to store the elements. For each operation, we would iterate through this list to find, insert, or remove elements, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1) on average for all operations.

What algorithm or data structure is used to solve Insert Delete GetRandom O(1) - Duplicates allowed?

Insert Delete GetRandom O(1) - Duplicates allowed uses the following concepts: Array, Hash Table, Math, Design, Randomized. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Insert Delete GetRandom O(1) - Duplicates allowed?

Explain your thought process clearly during the design phase. Discuss trade-offs between different data structures. Practice coding the solution without looking at references.

What are common mistakes when solving Insert Delete GetRandom O(1) - Duplicates allowed?

Not updating the hash map correctly after removals. Forgetting to handle duplicates properly.

#381

Insert Delete GetRandom O(1) - Duplicates allowed

Hard

ArrayHash TableMathDesignRandomizedHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1) on average for all operations
Space	O(1)	O(n)

💡

Intuition

Time O(1) on average for all operationsSpace O(n)

In the optimal approach, we use a combination of a list and a hash map. The list allows us to store elements and access them randomly, while the hash map keeps track of the indices of each element for efficient insertions and removals.

⚙️

Algorithm

3 steps

1Step 1: For insert, add the value to the list and update the hash map with the index of the new value.
2Step 2: For remove, find the index of the value in the hash map, swap it with the last element in the list, and remove the last element. Update the hash map accordingly.
3Step 3: For getRandom, simply return a random element from the list.

solution.py30 lines

1import random
2from collections import defaultdict
3
4class RandomizedCollection:
5    def __init__(self):
6        self.collection = []
7        self.index_map = defaultdict(set)
8
9    def insert(self, val):
10        self.collection.append(val)
11        self.index_map[val].add(len(self.collection) - 1)
12        return len(self.index_map[val]) == 1
13
14    def remove(self, val):
15        if val not in self.index_map or not self.index_map[val]:
16            return False
17        # Get an index of the element to remove
18        remove_idx = self.index_map[val].pop()
19        last_element = self.collection[-1]
20        self.collection[remove_idx] = last_element
21
22        # Update index_map for the last element
23        if self.index_map[last_element]:
24            self.index_map[last_element].remove(len(self.collection) - 1)
25            self.index_map[last_element].add(remove_idx)
26        self.collection.pop()
27        return True
28
29    def getRandom(self):
30        return random.choice(self.collection)

ℹ

Complexity note: The use of a list allows O(1) access for random elements, while the hash map allows O(1) average time complexity for insertions and removals by keeping track of indices.

1Using a hash map allows for efficient index tracking.
2Swapping elements during removal avoids shifting elements.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.