The Two Sneaky Numbers of Digitville

Problem Description

Welcome to Digitville, where numbers are sneaky little creatures! Imagine a town where every number from 1 to 100 is trying to hide from you. Your mission, should you choose to accept it, is to find the two numbers that have decided to show up twice at the local number party. Yes, you heard it right! Just like that one friend who always shows up uninvited, these numbers are crashing the party, and it’s your job to identify them.

So, how do you go about this? Well, you could try counting them like a bored teacher on a field trip, or you could use a more efficient method. Spoiler alert: the latter is what we’re going to do!

Code Solution


class Solution {
 public:
  vector getSneakyNumbers(const vector& nums) {
    constexpr int kMax = 100;
    vector ans;
    vector count(kMax + 1);

    for (const int num : nums)
      if (++count[num] == 2)
        ans.push_back(num);

    return ans;
  }
};

Approach

The approach here is as straightforward as counting sheep, but with a twist! We maintain a count of how many times each number appears in the input vector. When we find a number that appears for the second time, we add it to our answer list. This way, we efficiently track the sneaky numbers without losing our minds!

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n), where n is the number of elements in the input vector.
Space Complexity	O(1) if we consider the count array size as constant (since it only goes up to 100).

Real-World Example

Imagine you’re at a party with 100 guests, and you’re trying to figure out who brought the same dish twice. You could either taste every dish (which is time-consuming and risky for your waistline) or just keep a mental note of who brought what. In this case, the sneaky numbers are like those guests who thought they could sneak in a second batch of their famous potato salad without anyone noticing!