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 integernum
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 topopSmallest
andaddBack
.
Solution
Python3
class SmallestInfiniteSet:
def __init__(self):
self.A = list(range(1, 1001))
self.removed = set()
def popSmallest(self) -> int:
x = heappop(self.A)
self.removed.add(x)
return x
def addBack(self, num: int) -> None:
if num not in self.removed: return
self.removed.remove(num)
heappush(self.A, num)
# Your SmallestInfiniteSet object will be instantiated and called as such:
# obj = SmallestInfiniteSet()
# param_1 = obj.popSmallest()
# obj.addBack(num)