# 2336. Smallest Number in Infinite Set
###### tags: `leetcode`
## Description
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:

>$1 \leq num \leq 1000$
At most 1000 calls will be made in total to popSmallest and addBack.

## Solution
- Use intuitive way to achieve it is fine because it is the smallest element, so the difference would be be huge
- Iterate from the begining each time to find the smallest element

for (int i = 1; i <= 1000; i++)
{
    if (out[i])
    {
        out[i] = false;
        return i;
    }
}
return NULL;