How do you solve The Two Sneaky Numbers of Digitville on LeetCode?

The Two Sneaky Numbers of Digitville (LeetCode #3289) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap allows for efficient counting of occurrences.

What is the brute force solution for The Two Sneaky Numbers of Digitville?

The brute force approach for The Two Sneaky Numbers of Digitville is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each number in the list against every other number to count occurrences. This is straightforward but inefficient for larger lists. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve The Two Sneaky Numbers of Digitville?

The Two Sneaky Numbers of Digitville uses the following concepts: Array, Hash Table, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for The Two Sneaky Numbers of Digitville?

Clarify the problem statement before diving into coding. Think about edge cases and how your solution handles them. Explain your thought process as you code.

What are common mistakes when solving The Two Sneaky Numbers of Digitville?

Not checking for duplicates correctly. Overlooking the constraints of the problem.

#3289

The Two Sneaky Numbers of Digitville

Easy

ArrayHash TableMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages a HashMap to count occurrences efficiently. This allows us to find the sneaky numbers in a single pass through the list.

⚙️

Algorithm

3 steps

1Step 1: Create a HashMap to store the count of each number.
2Step 2: Traverse the list and update the count for each number in the HashMap.
3Step 3: After counting, iterate through the HashMap to find the two numbers that have a count of 2.

solution.py7 lines

1from collections import Counter
2
3def findSneakyNumbers(nums):
4    count = Counter(nums)
5    return [num for num, cnt in count.items() if cnt == 2]
6
7print(findSneakyNumbers([0, 1, 1, 0]))

ℹ

Complexity note: This complexity is linear because we make a single pass through the list to count occurrences, and then a second pass through the HashMap to find the sneaky numbers.

1Using a HashMap allows for efficient counting of occurrences.
2Understanding the problem constraints helps in choosing the right approach.

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