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Leetcode #2336: Smallest Number in Infinite Set

In this guide, we solve Leetcode #2336 Smallest Number in Infinite Set in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

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.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Design, Hash Table, Ordered Set, Heap (Priority Queue)

Intuition

Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.

By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.

Approach

Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.

This keeps the solution linear while remaining easy to explain in an interview setting.

Steps:

  • Initialize a hash map for seen items or counts.
  • Iterate through the input, querying/updating the map.
  • Return the first valid result or the final computed value.

Example

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.

Python Solution

class SmallestInfiniteSet: def __init__(self): self.s = SortedSet(range(1, 1001)) def popSmallest(self) -> int: x = self.s[0] self.s.remove(x) return x def addBack(self, num: int) -> None: self.s.add(num) # Your SmallestInfiniteSet object will be instantiated and called as such: # obj = SmallestInfiniteSet() # param_1 = obj.popSmallest() # obj.addBack(num)

Complexity

The time complexity is O(n×log⁡n)O(n \times \log n)O(n×logn). The space complexity is O(n)O(n)O(n).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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