How do you solve Smallest Number in Infinite Set on LeetCode?

Smallest Number in Infinite Set (LeetCode #2336) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Using a heap allows efficient retrieval of the smallest number.

What is the brute force solution for Smallest Number in Infinite Set?

The brute force approach for Smallest Number in Infinite Set is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves maintaining a simple list of popped numbers and always checking for the smallest number in the infinite set. This is straightforward but inefficient as it requires scanning through the list frequently. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Smallest Number in Infinite Set?

Think about how to efficiently manage state (like the smallest number). Discuss your thought process and any trade-offs in your approach. Practice implementing data structures like heaps and sets.

What are common mistakes when solving Smallest Number in Infinite Set?

Not using a heap for efficient retrieval. Failing to check if a number is already in the set before adding it back.

#2336

Smallest Number in Infinite Set

Medium

Hash TableDesignHeap (Priority Queue)Ordered SetHeapPriority Queue

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(n)	O(n)

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Intuition

Time O(log n)Space O(n)

The optimal approach uses a priority queue (min-heap) to efficiently manage the smallest available numbers and a set to track which numbers have been added back. This allows us to quickly pop the smallest number and manage additions without scanning through a list.

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Algorithm

4 steps

1Step 1: Use a min-heap to store numbers that have been added back.
2Step 2: Maintain a variable for the current smallest number to be popped.
3Step 3: For popSmallest(), check the heap for the smallest number, or return the current smallest if the heap is empty, then increment the current smallest.
4Step 4: For addBack(num), add num to the heap if it's less than the current smallest.

solution.py16 lines

1import heapq
2class SmallestInfiniteSet:
3    def __init__(self):
4        self.min_heap = []
5        self.current = 1
6
7    def popSmallest(self):
8        if self.min_heap:
9            return heapq.heappop(self.min_heap)
10        smallest = self.current
11        self.current += 1
12        return smallest
13
14    def addBack(self, num):
15        if num < self.current:
16            heapq.heappush(self.min_heap, num)

ℹ

Complexity note: The time complexity is O(log n) for each popSmallest and addBack operation due to the heap operations, making it efficient compared to the brute force approach.

1Using a heap allows efficient retrieval of the smallest number.
2Tracking the current smallest number helps manage additions effectively.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.