Problem description:

Design a data structure that supports all following operations in average O(1) time.

insert(val): Inserts an item val to the set if not already present.
remove(val): Removes an item val from the set if present.
getRandom: Returns a random element from current set of elements. Each element must have the same probability of being returned.
Example:

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// Init an empty set.
RandomizedSet randomSet = new RandomizedSet();

// Inserts 1 to the set. Returns true as 1 was inserted successfully.
randomSet.insert(1);

// Returns false as 2 does not exist in the set.
randomSet.remove(2);

// Inserts 2 to the set, returns true. Set now contains [1,2].
randomSet.insert(2);

// getRandom should return either 1 or 2 randomly.
randomSet.getRandom();

// Removes 1 from the set, returns true. Set now contains [2].
randomSet.remove(1);

// 2 was already in the set, so return false.
randomSet.insert(2);

// Since 2 is the only number in the set, getRandom always return 2.
randomSet.getRandom();

Solution:

In this question, we need to randomly pick a element from the inserted numbers. To do so, we need to know two things:

  1. The numbers we currently have
  2. Where are they(index)

We use a dictionary to store the value-index mapping, an array to store all the numbers.

Also, the question mentioned want to make each operation O(1). We have two choice to delete the element in array:

  1. array.remove(val): takes O(n) time
  2. move the element to last index then array.pop(): O(1)
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class RandomizedSet:

def __init__(self):
"""
Initialize your data structure here.
"""
self.dic, self.nums = {}, []

def insert(self, val: int) -> bool:
"""
Inserts a value to the set. Returns true if the set did not already contain the specified element.
"""
if val in self.dic:
return False
else:
self.nums.append(val)
self.dic[val] = len(self.nums)-1
return True

def remove(self, val: int) -> bool:
"""
Removes a value from the set. Returns true if the set contained the specified element.
"""
if val in self.dic:
# move the last index value to where val is
# then pop the last element in array to achieve O(1) remove
idx = self.dic[val]
last = self.nums[-1]
last_idx = self.dic[last]
self.nums[idx] = last
self.dic[last] = idx
self.nums.pop()
del self.dic[val]

return True
else:
return False


def getRandom(self) -> int:
"""
Get a random element from the set.
"""
return random.choice(self.nums)
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class RandomizedSet {
public:
//store input numbers in vector
//save index and
unordered_map<int, int> m;
vector<int> nums;

/** Initialize your data structure here. */
RandomizedSet() {

}

/** Inserts a value to the set. Returns true if the set did not already contain the specified element. */
bool insert(int val) {
if (m.find(val) != m.end()) return false;
nums.push_back(val);
m[val] = nums.size() - 1;
return true;
}

/** Removes a value from the set. Returns true if the set contained the specified element. */
bool remove(int val) {
if (m.find(val) == m.end()) return false;
int last = nums.back();
m[last] = m[val];
nums[m[val]] = last;
nums.pop_back();
m.erase(val);
return true;
}

/** Get a random element from the set. */
int getRandom() {
int idx= rand()%nums.size();
return nums[idx];
}
};

/**
* Your RandomizedSet object will be instantiated and called as such:
* RandomizedSet obj = new RandomizedSet();
* bool param_1 = obj.insert(val);
* bool param_2 = obj.remove(val);
* int param_3 = obj.getRandom();
*/

time complexity: $O(1)$
space complexity: $O(n)$
reference: