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Java-based LeetCode algorithm problem solutions, regularly updated
2336\. Smallest Number in Infinite Set
Medium
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`.