3289. The Two Sneaky Numbers of Digitville

EasyArrayHash TableMath

Problem Description

In the town of Digitville, there is a list of numbers called nums containing integers from 0 to n - 1. Each number is supposed to appear exactly once in the list. However, two mischievous numbers sneaked in, each appearing an additional time, making the list longer than usual. 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 to restore the original state of peace in Digitville.

Intuition

To identify the two sneaky numbers, we can utilize a simple counting technique. By using a data structure that can count the frequency of each number, we traverse through the given list nums. While traversing, we keep track of how many times each number appears. The numbers that appear more than once in the list will have a count of two. By collecting these numbers, we can construct the required pair of sneaky numbers. This approach efficiently allows us to solve the problem by leveraging the ability to count occurrences, making it both intuitive and straightforward to implement.

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Solution Approach

The solution is built around a counting mechanism to track the frequency of each number in the list nums. The Counter class from the collections module is well-suited for this task. Here’s how the solution is implemented:

Counting Frequency: We first create an instance of Counter using the list nums. This effectively creates a dictionary-like structure where each number from nums is a key, and its value is the count of occurrences of that number in the list.
```
cnt = Counter(nums)
```
Finding Sneaky Numbers: With the counts established, we can easily identify the numbers that appear twice. We iterate over the items in cnt and collect those numbers whose count is exactly two. This gives us our sneaky numbers.
```
return [x for x, v in cnt.items() if v == 2]
```
Outputting the Result: The sneaky numbers are stored in a list and returned as output. The order of the numbers in the returned list is not specified, as the problem statement allows them to be in any order.

This approach is intuitive as it directly uses counting to identify the problematic duplicate numbers. It efficiently utilizes the Counter to perform counting in a single pass over nums, while a subsequent pass identifies the needed numbers, yielding a linear time complexity overall.

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Example Walkthrough

Let's walk through an example to understand the solution approach.

Consider the list nums = [0, 1, 2, 3, 2, 4, 5, 3]. This list contains numbers from 0 to 5, but two numbers appear twice due to the mischievous acts we need to uncover. Here's how we can solve the problem using the described approach:

Counting Frequency:
- We use the Counter class from the collections module to count how many times each number appears in the list nums.
- After initializing Counter with our list, we have the following counts:
```
cnt = Counter(nums) # Result: {0: 1, 1: 1, 2: 2, 3: 2, 4: 1, 5: 1}
```
- Here, the numbers 2 and 3 appear twice, which indicates they are the sneaky numbers.

Finding Sneaky Numbers:

We iterate over the items in cnt and check which numbers have a frequency of 2.

Construct a list of such numbers:

sneaky_numbers = [x for x, v in cnt.items() if v == 2] # Result: [2, 3]

Outputting the Result:
- The list [2, 3] is returned as the result. The order does not matter, so either [2, 3] or [3, 2] would be correct.

Thus, the two sneaky numbers causing mischief in the town of Digitville are identified as 2 and 3.

Solution Implementation

1from typing import List
2from collections import Counter
3
4class Solution:
5    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
6        # Count the occurrences of each number in the input list using Counter
7        count = Counter(nums)
8      
9        # Collect numbers that appear exactly twice in the list
10        sneaky_numbers = [number for number, frequency in count.items() if frequency == 2]
11      
12        return sneaky_numbers
13

1class Solution {
2    public int[] getSneakyNumbers(int[] nums) {
3        // Create an array to hold the two numbers that appear twice
4        int[] result = new int[2];
5      
6        // Create a counter array to keep track of the frequency of each number
7        int[] count = new int[100];
8      
9        // Index to store the numbers that appear exactly twice
10        int index = 0;
11      
12        // Iterate over each number in the input array
13        for (int number : nums) {
14            // Increase the count of the current number
15            count[number]++;
16          
17            // If the count of the current number becomes 2, add to result
18            if (count[number] == 2) {
19                result[index++] = number;
20            }
21        }
22      
23        // Return the result array containing numbers that appeared twice
24        return result;
25    }
26}
27

1#include <vector>
2
3class Solution {
4public:
5    // This function returns a vector of numbers that appear at least twice in the input vector nums.
6    std::vector<int> getSneakyNumbers(std::vector<int>& nums) {
7        std::vector<int> result;
8        int count[100] = {}; // Initialize an array to count occurrences of each number in the range [0, 99].
9      
10        // Iterate over each number in nums
11        for (int number : nums) {
12            // Increment the count for this number and check if it's the second occurrence.
13            if (++count[number] == 2) {
14                // If this is the second occurrence, add the number to the result vector.
15                result.push_back(number);
16            }
17        }
18      
19        return result; // Return the list of numbers that appear at least twice.
20    }
21};
22

1// Function to identify "sneaky" numbers in an array, where "sneaky" numbers are those
2// that appear more than once.
3function getSneakyNumbers(nums: number[]): number[] {
4    // Initialize an array to store the result.
5    const ans: number[] = [];
6
7    // Initialize an array of size 100 to count occurrences of numbers.
8    // Assuming numbers in 'nums' range from 0 to 99.
9    const cnt: number[] = Array(100).fill(0);
10
11    // Iterate through each number in the input array.
12    for (const x of nums) {
13        // Increment the count for each number.
14        cnt[x]++;
15
16        // If a number has been seen more than once, it's a sneaky number; add it to the result.
17        if (cnt[x] > 1) {
18            ans.push(x);
19        }
20    }
21
22    // Return the array of sneaky numbers.
23    return ans;
24}
25

Time and Space Complexity

The given code has a time complexity of O(n), where n is the length of the array nums. This is because creating the Counter object requires iterating through the list once, resulting in a linear time operation. Additionally, generating the list of sneaky numbers by iterating over the items in the Counter object also requires linear time, although typically less than n since not every element will be duplicated exactly twice.

The space complexity of the code is also O(n). This is due to the space needed to store information in the Counter object, which at most holds unique elements of the list and their counts. Similarly, the output list that stores the numbers occurring exactly twice will also require space proportional to n in the worst-case scenario.

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Question 1 out of 10

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12

1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18

1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

1 2 3 6 x x

1 2 3 x x x 6 x x

1 2 6 x x x 3 x x

I don’t know