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The Two Sneaky Numbers of Digitville

Number: 3581

Difficulty: Easy

Paid? No

Companies: N/A

Problem Description

In Digitville, you are given an array nums containing integers from 0 to n - 1 where each number is supposed to appear exactly once. However, two numbers appear twice. Your task is to identify these two repeated numbers and return them in any order.

Key Insights

  • The size of the array is n + 2; exactly two numbers are repeated.
  • The numbers are constrained between 0 and n - 1.
  • We can count the occurrence of each number using a hash table.
  • The problem can be solved in linear time.
  • Simplicity of the constraints allows a straightforward frequency counting approach.

Space and Time Complexity

Time Complexity: O(n) - We traverse the array once. Space Complexity: O(n) - In the worst case, a hash table is used to store counts for up to n numbers.

Solution

The solution uses a hash table (or dictionary) to count the frequency of each number in the array. By iterating over the array once, we update the count for each number. Finally, we iterate through the hash table and select the numbers that have a count of exactly two. This approach directly leverages the guarantee of exactly two repeated numbers and provides a clear and efficient solution.

Code Solutions

# Python solution using a dictionary for frequency counting.

def findSneakyNumbers(nums):
    # Dictionary to store the frequency of each number
    frequency = {}
    # Iterate over every number in the array
    for num in nums:
        # Increment count in dictionary; if num not in dictionary, get default 0
        frequency[num] = frequency.get(num, 0) + 1
    # List to store the two sneaky numbers that appear twice
    result = []
    # Check each key in the dictionary for frequency equal to 2
    for num, count in frequency.items():
        if count == 2:
            result.append(num)
    return result

# Example usage:
# print(findSneakyNumbers([0, 3, 2, 1, 3, 2]))  # Output: [3, 2]
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