The Two Sneaky Numbers of Digitville

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Code Solution


class Solution {
  public int[] getSneakyNumbers(int[] nums) {
    final int kMax = 100;
    int[] ans = new int[2];
    int[] count = new int[kMax + 1];
    int ansIndex = 0;

    for (final int num : nums)
      if (++count[num] == 2)
        ans[ansIndex++] = num;

    return ans;
  }
}

Problem Description

Welcome to the whimsical world of Digitville, where numbers are sneaky little creatures that love to hide! The problem at hand is to find two numbers that have decided to play hide-and-seek in a list. Imagine you’re at a party, and two of your friends keep sneaking off to the snack table for a second helping of chips. Your job is to identify these two sneaky numbers from a given array of integers, where each number can appear at most twice.

So, if you have a list like [1, 2, 3, 2, 1, 4], you should be able to spot that 1 and 2 are the sneaky culprits who just couldn’t resist a second round of snacks!

Approach

The code uses a simple counting mechanism to track how many times each number appears in the input array. It maintains a count array of size 101 (to accommodate numbers from 0 to 100). As it iterates through the input array, it increments the count for each number. When a number’s count reaches 2, it gets added to the result array. This way, we efficiently find the two sneaky numbers without any unnecessary complexity.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n)
Space Complexity	O(1)

Real-World Example

Imagine you’re at a family reunion, and everyone is trying to sneak extra servings of grandma’s famous pie. You notice that Uncle Bob and Cousin Sally keep going back for more. Just like in our problem, you can easily identify them by counting how many times they’ve been spotted at the dessert table. In the end, you can confidently declare them the “Sneaky Pie Lovers of the Reunion!”