2336. Smallest Number in Infinite Set

#### QUESTION:

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.
``````

#### SOLUTION:

``````class SmallestInfiniteSet() {

var arr:BooleanArray = BooleanArray(1001 ){ i -> true}

init {
arr[0] = false
}

fun popSmallest(): Int {
var tmp:Int = 0
while (!arr[tmp]) {
tmp++
}
arr[tmp] = false
return tmp
}

fun addBack(num: Int) {
arr[num] = true
}

}

``````
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