# Problem 3289: The Two Sneaky Numbers of Digitville

## Problem Statement

In the town of Digitville, there was a list of numbers called `nums` containing integers from 0 to `n - 1`. Each number was supposed to appear exactly once in the list, however, two mischievous numbers sneaked in an additional time, making the list longer than usual.

## Task

As the town detective, your task is to find these two sneaky numbers. Return an array of size two containing the two numbers (in any order), so peace can return to Digitville.

### Difficulty

* Easy

### Examples

#### Example 1
* **Input:** `nums = [0,1,1,0]`
* **Output:** `[0,1]`
* **Explanation:** The numbers 0 and 1 each appear twice in the array.

#### Example 2
* **Input:** `nums = [0,3,2,1,3,2]`
* **Output:** `[2,3]`
* **Explanation:** The numbers 2 and 3 each appear twice in the array.

#### Example 3
* **Input:** `nums = [7,1,5,4,3,4,6,0,9,5,8,2]`
* **Output:** `[4,5]`
* **Explanation:** The numbers 4 and 5 each appear twice in the array.

### Notes

* The input list contains integers from 0 to `n - 1`, where `n` is the length of the input list. Two numbers appear twice in the list, and the rest appear once.

Q-can you explain it using your own words ? 
Oky so we are given a list of integers , the common rules that each number in this list most appear at most 1 times 
But 2 malicious integers  does not respect this rule … 
You task is to find them and return theme as list 
Q- what is the input/output format ? 
Input : List[int] 
Output : List[int] 

Q- can you go through examples ? 

Input : [0,3,2,1,3,2] 

The rule tell use that  the numbers within this list 
So looking at the list we see that only 3 and 2 are appears more then one time 
So we can return [2,3] as a result to this example 

Q- do have any technique , pattern , particul ds to solve this ? 
I think this problem can be solving by two steps 
Step 1 : we need to sort the list to facilte the search for duplicate element 
Step 2 : search the elements that appear more than one time by computing the two adjacents ones 

Q- can you more formal pesudoCode ? 

Functions : def getSneakyNumbers(self, nums: List[int]) -> List[int]:
		// Q- why we need to compute the length of the list 
		// A- it help us traverse the list more easily and make the code more readable 
		n=nums.length() 
		// Q- why we need to sort the input list 
		// A- it facilitate the detection of duplicate element , by traverse the list paiwize with adjancent values 
		nums.sort() 
		// Q- why we need to initiate a an empty list 
		// A- it help us store the anomalies values 
		result=[] 
		// Q- why we use for loop not while loop 
		// A- because the problem is more iteration problem than traversal one , and also the length of the input collection is already known (computed )
		For i from 1 to n 
		// Q- why you start from the second element 
		// A- it a choice not mandatory but the like the pattern current & prev to check the duplication problem in a collection 
		If nums[i]==nums[i-1]: 
			// Q-why we you add it to the list when the if condition is true 
			// A- because if two adjance elements are equal we can infer that the element are anomaly
			result.append(nums[i]) 
		Return result 

Q- now you are a mentor to me you do not give the directed solution 
First you explain the problem using formal story telling technique 
Then in the storytelling you go through the example step by step 
Then you give a feeadbakc about my approach 
Then you give give some hints or guidness oky to approch the full solution 
Let's break down this detective story in Digitville. We have a list of numbers, nums, supposedly containing each number from 0 up to n-1 exactly once. Think of it like a town census where every resident has a unique ID. But, two sneaky numbers have duplicated themselves in the list, throwing off the count. Our job is to find these two impostors.
Let's walk through Example 2: nums = [0, 3, 2, 1, 3, 2]. The "census" should only have numbers from 0 to 5 (because the length is 6), each appearing once. But if we look closely, we see both 2 and 3 appear twice. So, our answer should be [2, 3] (or [3, 2], the order doesn't matter).
Your approach of sorting the list and then comparing adjacent elements is excellent! It leverages a fundamental pattern in algorithm design: sorting simplifies many problems related to searching and finding duplicates.
Here's some feedback and guidance on your pseudocode:
Length Calculation: You correctly identified the need to calculate the length. This provides a clear stopping point for our iteration.
Sorting: Sorting is key here. It brings duplicate numbers together, making them easy to detect in a single pass.
Result List Initialization: Spot on! We need a place to store our discovered sneaky numbers.
Looping: You're right to use a for loop. We know exactly how many times we need to iterate (n-1 comparisons), which makes a for loop the natural choice.
Starting from the Second Element: You're thinking ahead! By starting at the second element (index 1), you can compare it directly to the previous element. This avoids going out of bounds.
Here's a slightly refined pseudocode, incorporating these points and adding some clarity:
     function getSneakyNumbers(nums):
  n = length(nums)
  sort(nums) // Sort the input list in place
  result = [] // Initialize an empty list to store the sneaky numbers

  for i from 1 to n - 1:  // Iterate from the second element to the end
    if nums[i] == nums[i-1]: // Compare the current element with the previous
      result.append(nums[i])  // If they are the same, it's a sneaky number

  return result
   
content_copy Use code with caution.
This pseudocode explicitly mentions sorting the list in place and clearly defines the loop boundaries. It also emphasizes the logic: comparing each element with its predecessor to detect duplicates. You're on the right track, and this refined pseudocode provides a solid foundation for implementing a working solution. Now you can translate this into your preferred programming language! Remember to handle potential edge cases (like empty input lists) in your final implementation.

class Solution:
    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
        n=len(nums)
        nums.sort()
        result=[]
        for i in range(1,n):
            if nums[i]==nums[i-1]:
                result.append(nums[i])
        return result
       

I solve it and i gotte a code “TimeComplexity ‘ Space but what about “Sapce compxity” 

How tho approach this better 
Mentor's Guidance: Unmasking the Sneaky Numbers of Digitville
Understanding the Problem: A Story from Digitville
Imagine Digitville, a town where every citizen has a unique ID from 0 to n-1, like a perfectly ordered census. But two mischievous numbers have duplicated themselves, throwing the registry into chaos. Your task, as Digitville's detective, is to find these impostors.
Walking Through an Example: The Case of Numbers 2 and 3
Consider nums = [0, 3, 2, 1, 3, 2]. Digitville expects numbers 0 to 5, each appearing once. However, both 2 and 3 show up twice. Therefore, the sneaky numbers are [2, 3].
Feedback on Your Approach
Sorting and comparing adjacent elements is a good start. Sorting groups duplicates together, simplifying their detection. However, let's refine this further:
Edge Cases: While sorting helps, ensure your solution handles cases where duplicates are at the beginning or end of the list.
Efficiency: Sorting typically takes O(n log n) time. Can we do better? Strive for O(n) time and O(1) space – a single pass and no extra data structures.
Hints to Guide Your Investigation
Leverage the Range: The numbers are from 0 to n-1. How can you use this to track occurrences without extra space? Think about using the input array itself to store information.
Index Mapping: Consider using each number as an index into the nums array. Modifying the value at that index can signal an element's presence. Think about how you can mark an element as "seen" without losing the original value completely.
In-Place Detection: Can you detect duplicates during the traversal, avoiding a second pass? Think about arithmetic operations that can help you differentiate between a first and second encounter of a number.
By considering these hints, you can devise a clever and efficient solution to restore order in Digitville. Good luck, detective!