Leetcode #380: Insert Delete GetRandom O(1)
In this guide, we solve Leetcode #380 Insert Delete GetRandom O(1) in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Implement the RandomizedSet class: RandomizedSet() Initializes the RandomizedSet object. bool insert(int val) Inserts an item val into the set if not present.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Array, Hash Table, Math, Randomized
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class RandomizedSet:
def __init__(self):
self.d = {}
self.q = []
def insert(self, val: int) -> bool:
if val in self.d:
return False
self.d[val] = len(self.q)
self.q.append(val)
return True
def remove(self, val: int) -> bool:
if val not in self.d:
return False
i = self.d[val]
self.d[self.q[-1]] = i
self.q[i] = self.q[-1]
self.q.pop()
self.d.pop(val)
return True
def getRandom(self) -> int:
return choice(self.q)
# Your RandomizedSet object will be instantiated and called as such:
# obj = RandomizedSet()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.