Watch 10 video solutions for The Two Sneaky Numbers of Digitville, a easy level problem involving Array, Hash Table, Math. This walkthrough by codestorywithMIK has 4,352 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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.
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.
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.
Constraints:
2 <= n <= 100nums.length == n + 20 <= nums[i] < nnums contains exactly two repeated elements.Problem Overview: You receive an integer array that contains numbers from 0 to n-1, but two numbers appear twice. Every other number appears exactly once. The task is to identify those two duplicated values (the “sneaky numbers”) and return them.
Approach 1: Frequency Dictionary (Hash Table) (Time: O(n), Space: O(n))
The most direct solution uses a frequency counter. Iterate through the array and store counts in a hash table (dictionary). Each time you encounter a number, increment its frequency. When a value reaches a count of 2, it is one of the sneaky numbers. Continue scanning until both duplicates are found. Hash tables provide average O(1) insert and lookup operations, so the entire array can be processed in linear time.
This approach works well for unsorted arrays and is typically the first solution engineers reach for when detecting duplicates. It trades additional memory for simplicity and speed. The technique relies on efficient key lookups from a hash table while iterating once through the array. If interview constraints allow extra space, this is the cleanest implementation.
Approach 2: Sorting and Pairwise Comparison (Time: O(n log n), Space: O(1) or O(log n))
A second approach avoids explicit extra storage by sorting the array first. Once sorted, duplicate values will appear next to each other. After sorting, iterate through the array and compare each element with the next one. Whenever nums[i] == nums[i+1], that value is one of the sneaky numbers.
The key insight is that sorting groups identical values together, turning duplicate detection into a simple adjacency check. The main cost comes from the sorting step, which takes O(n log n) time. Depending on the language and sorting implementation, space usage ranges from constant to O(log n). This approach is useful when memory is constrained or when the array is already nearly sorted.
Recommended for interviews: The hash table solution is usually the expected answer because it achieves optimal O(n) time with straightforward logic. Interviewers often accept the sorting approach as a valid alternative, especially if you discuss the tradeoff between extra memory and sorting overhead. Demonstrating both shows strong understanding of hash table lookup patterns and fundamental counting strategies used for duplicate detection.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Frequency Dictionary (Hash Table) | O(n) | O(n) | Best general solution for unsorted arrays when extra memory is allowed |
| Sorting + Pairwise Comparison | O(n log n) | O(1) to O(log n) | Useful when minimizing extra space or when the array can be sorted |