# Solving Leetcode Interviews in Seconds with AI: Smallest Number in Infinite Set


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2336" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You have a set which contains all positive integers [1, 2, 3, 4, 5, ...]. Implement the SmallestInfiniteSet class:  SmallestInfiniteSet() Initializes the SmallestInfiniteSet object to contain all positive integers. int popSmallest() Removes and returns the smallest integer contained in the infinite set. void addBack(int num) Adds a positive integer num back into the infinite set, if it is not already in the infinite set.    Example 1:  Input ["SmallestInfiniteSet", "addBack", "popSmallest", "popSmallest", "popSmallest", "addBack", "popSmallest", "popSmallest", "popSmallest"] [[], [2], [], [], [], [1], [], [], []] Output [null, null, 1, 2, 3, null, 1, 4, 5]  Explanation SmallestInfiniteSet smallestInfiniteSet = new SmallestInfiniteSet(); smallestInfiniteSet.addBack(2);    // 2 is already in the set, so no change is made. smallestInfiniteSet.popSmallest(); // return 1, since 1 is the smallest number, and remove it from the set. smallestInfiniteSet.popSmallest(); // return 2, and remove it from the set. smallestInfiniteSet.popSmallest(); // return 3, and remove it from the set. smallestInfiniteSet.addBack(1);    // 1 is added back to the set. smallestInfiniteSet.popSmallest(); // return 1, since 1 was added back to the set and                                    // is the smallest number, and remove it from the set. smallestInfiniteSet.popSmallest(); // return 4, and remove it from the set. smallestInfiniteSet.popSmallest(); // return 5, and remove it from the set.    Constraints:  1 <= num <= 1000 At most 1000 calls will be made in total to popSmallest and addBack.  

	# Explanation
	*   **Maintain a sorted set of added-back numbers:** Use a min-heap (priority queue) to efficiently store numbers that have been added back into the set after being popped. This ensures quick access to the smallest available number.
*   **Track the smallest "missing" number:** Keep track of the smallest positive integer that has not yet been popped. This avoids unnecessary heap operations when popping the smallest element in the infinite set.

*   **Runtime Complexity:** O(log K) for `addBack` and `popSmallest`, where K is the number of calls to `addBack`. O(1) for initialization.
    **Storage Complexity:** O(K), where K is the number of calls to `addBack`.

	
	# Code
	```python
	import heapq

class SmallestInfiniteSet:

    def __init__(self):
        self.heap = []
        self.smallest = 1

    def popSmallest(self):
        if self.heap:
            smallest_added = heapq.heappop(self.heap)
            return smallest_added
        else:
            self.smallest += 1
            return self.smallest - 1

    def addBack(self, num):
        if num < self.smallest and num not in self.heap:
            heapq.heappush(self.heap, num)


# Your SmallestInfiniteSet object will be instantiated and called as such:
# obj = SmallestInfiniteSet()
# param_1 = obj.popSmallest()
# obj.addBack(num)
	```