#2336

Medium

LeetCode Problem Workspace

Smallest Number in Infinite Set

Design a data structure to handle the smallest missing element in an infinite set, with the ability to add and remove elements efficiently.

hash table design heap priority queue ordered set

Problem Statement

You are tasked with implementing a data structure that handles an infinite set of positive integers, starting from 1, with the ability to pop the smallest number and add back previously popped numbers. The set will be manipulated through two key operations: popSmallest, which removes and returns the smallest number, and addBack, which adds a number back into the set if it has been previously popped.

To efficiently implement this data structure, you need to ensure that operations like popSmallest and addBack are optimized for time and space. The design should handle adding back numbers that were previously removed and ensure that the smallest available number can always be identified quickly. Given the constraints, you should aim for a solution that avoids brute force methods, relying instead on optimized data structures like heaps and hash tables.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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 Approach

Use of Hash Table

A hash table will be used to track numbers that have been added back into the set. This allows quick lookups to check if a number is available to be popped back. The hash table ensures that duplicates are not inserted and supports efficient O(1) operations for checking if a number is in the set.

Heap for Tracking the Smallest Element

A heap (priority queue) will be used to efficiently track the smallest element that is available to be popped. The heap allows for fast retrieval and removal of the smallest element, providing O(log n) complexity for insertion and removal operations.

Efficient Pop and AddBack Operations

The popSmallest operation will first check the heap for the smallest available number. If the number has been previously popped and added back, the system will pop it and update the heap. For addBack, the number is added to the hash table and pushed into the heap if it hasn't been reinserted already.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for popSmallest is O(log n) due to heap operations. The time complexity for addBack is O(log n) as well, for inserting into the heap. The space complexity is O(n), where n is the number of operations performed, since both the heap and the hash table store elements added back to the set.

What Interviewers Usually Probe

Candidates may struggle with balancing efficient popSmallest and addBack operations.
Look for solutions using a combination of data structures, especially heaps and hash tables.
Ensure the candidate understands the trade-off between space complexity and operation speed.

Common Pitfalls or Variants

Common pitfalls

Overusing brute force methods for tracking elements will lead to inefficiency.
Not handling the edge case where a number is popped and then added back multiple times.
Failing to optimize for time complexity in popSmallest or addBack, leading to suboptimal performance.

Follow-up variants

Handling a larger range of numbers efficiently, where the set grows even more.
Adding additional operations, such as checking if a specific number exists in the set.
Optimizing for worst-case time complexity when the number of operations increases.

FAQ

How can I efficiently solve the Smallest Number in Infinite Set problem?

To solve this problem efficiently, use a combination of a hash table for tracking added back elements and a heap for managing the smallest number.

What is the best way to optimize the popSmallest operation?

Using a heap for the popSmallest operation ensures that the smallest element is always accessible in O(log n) time.

What should I keep in mind while implementing addBack in this problem?

When implementing addBack, make sure to track the number in a hash table and add it back into the heap only if it hasn’t been reinserted already.

What are the time and space complexities of the solution?

The time complexity for both popSmallest and addBack is O(log n), while the space complexity is O(n), where n is the number of operations performed.

How does this problem relate to other heap-based problems?

This problem shares similarities with other heap-based problems, where you need to track and manipulate the smallest or largest element in a set efficiently.

terminal

Solution

Solution 1: Ordered Set + Simulation

We note that the range of elements in the set given by the problem is $[1, 1000]$, and the operations we need to support are:

1

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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)

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