#3289

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

The Two Sneaky Numbers of Digitville

Find the two numbers that appear twice in a list of integers from 0 to n-1 using array scanning plus hash lookup efficiently.

array hash table math

Problem Statement

In Digitville, a list called nums contains integers from 0 to n-1, but two numbers have been duplicated, making the list n+2 elements long. Each number should appear once, but two numbers appear twice.

As the town detective, identify the two sneaky numbers and return them in any order. For example, given nums = [0,1,1,0], the output should be [0,1] because both 0 and 1 appear twice.

Examples

Example 1

Input: nums = [0,1,1,0]

Output: [0,1]

The numbers 0 and 1 each appear twice in the array.

Example 2

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

Output: [2,3]

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]

The numbers 4 and 5 each appear twice in the array.

Constraints

2 <= n <= 100
nums.length == n + 2
0 <= nums[i] < n
The input is generated such that nums contains exactly two repeated elements.

Solution Approach

Hash Set Scan

Initialize an empty hash set and iterate through the array. Add each number to the set. If a number is already present, add it to the result list. This approach ensures O(n) time and handles duplicates efficiently.

Frequency Array

Create an auxiliary array of size n initialized to zero. Iterate through nums, incrementing the count for each number. Numbers with count greater than one are the sneaky numbers. This trades O(n) space for simpler counting logic.

Mathematical Difference

Compute the sum of 0 to n-1 and compare it with the sum of nums to find the total duplicate sum. Combine this with sum of squares to isolate individual repeated numbers. This approach avoids extra hash structures but requires careful algebra.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) for scanning the array once. Space complexity is O(n) for hash set or frequency array, though the mathematical approach reduces space to O(1). The main trade-off is simplicity versus extra space usage.

What Interviewers Usually Probe

Candidate recognizes using a hash table for quick duplicate detection.
Candidate attempts in-place or mathematical approaches to reduce space usage.
Candidate carefully handles index ranges to avoid off-by-one errors in the array.

Common Pitfalls or Variants

Common pitfalls

Failing to handle all duplicates when scanning, recording only the first repeated number.
Confusing indices with values when constructing frequency arrays.
Attempting sum-based approaches without considering two duplicates, leading to incorrect algebra.

Follow-up variants

Find k repeated numbers instead of two, generalizing the hash scan or frequency array.
Input may contain numbers outside 0 to n-1 range, requiring mapping to handle duplicates.
Return duplicates in sorted order, emphasizing order handling on top of detection.

FAQ

What is the simplest way to find the two sneaky numbers of Digitville?

Use a hash set to scan through nums and collect numbers that appear a second time. This directly applies the array scanning plus hash lookup pattern.

Can this problem be solved without extra space?

Yes, by using sum and sum of squares calculations to isolate the two duplicates, but it requires careful algebra.

Does the order of returned numbers matter?

No, the problem allows returning the two sneaky numbers in any order.

What common mistakes should I avoid?

Avoid recording only one duplicate, misindexing when counting frequencies, or incorrectly applying sum-based formulas for two numbers.

Why is this problem classified under Array, Hash Table, and Math?

It involves scanning arrays, using hash tables to track seen numbers, and optionally applying mathematical sums to detect duplicates efficiently.

terminal

Solution

Solution 1: Counting

We can use an array $\textit{cnt}$ to record the number of occurrences of each number.

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class Solution:
    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
        cnt = Counter(nums)
        return [x for x, v in cnt.items() if v == 2]

Solution 2: Bit Manipulation

Let the length of array $\textit{nums}$ be $n + 2$, which contains integers from $0$ to $n - 1$, with two numbers appearing twice.

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class Solution:
    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
        cnt = Counter(nums)
        return [x for x, v in cnt.items() if v == 2]

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