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

Insert Delete GetRandom O(1) (LeetCode #380) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Using a hash map allows for O(1) average time complexity for lookups.

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

The brute force approach for Insert Delete GetRandom O(1) is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach uses a simple list to store elements and relies on linear searches to find, insert, and remove elements. This is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(1).

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

Insert Delete GetRandom O(1) 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)?

Explain your thought process clearly as you code. Consider edge cases, such as inserting duplicates or removing non-existent elements. Practice writing tests for your implementation.

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

Not updating the index of the last element after removal. Confusing the order of operations when inserting and removing.

#380

Insert Delete GetRandom O(1)

Medium

ArrayHash TableMathDesignRandomizedHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n)	O(n)

💡

Intuition

Time O(1)Space O(n)

The optimal solution uses a combination of a hash map and an array to achieve O(1) time complexity for all operations. The hash map allows for quick lookups, while the array enables efficient random access.

⚙️

Algorithm

5 steps

1Step 1: Create a hash map to store the value and its index in the array.
2Step 2: Create an array to store the elements.
3Step 3: For insert, check if the value exists in the hash map; if not, add it to the array and hash map and return true; otherwise, return false.
4Step 4: For remove, find the index of the value in the array using the hash map, swap it with the last element, remove the last element, and update the hash map accordingly.
5Step 5: For getRandom, return a random element from the array.

solution.py23 lines

1import random
2class RandomizedSet:
3    def __init__(self):
4        self.map = {}
5        self.nums = []
6    def insert(self, val):
7        if val not in self.map:
8            self.map[val] = len(self.nums)
9            self.nums.append(val)
10            return True
11        return False
12    def remove(self, val):
13        if val in self.map:
14            last_element = self.nums[-1]
15            idx = self.map[val]
16            self.nums[idx] = last_element
17            self.map[last_element] = idx
18            self.nums.pop()
19            del self.map[val]
20            return True
21        return False
22    def getRandom(self):
23        return random.choice(self.nums)

ℹ

Complexity note: The time complexity is O(1) for all operations because the hash map allows for constant time lookups, and the array allows for constant time access to elements.

1Using a hash map allows for O(1) average time complexity for lookups.
2Swapping elements during removal helps maintain O(1) complexity.

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