The Two Sneaky Numbers of Digitville - Practice Coding Problems

The Two Sneaky Numbers of Digitville - Problem

Welcome to Digitville, a peaceful town where order and precision reign supreme! 🏘️

In this mathematical town, there was supposed to be a perfect list called nums containing exactly the integers from 0 to n-1, with each number appearing exactly once. However, two mischievous numbers decided to sneak back into the list for a second appearance, disrupting the town's harmony!

As the town's detective 🕵️, your mission is to identify these two sneaky duplicate numbers that have caused the list to become longer than its intended size of n.

Goal: Return an array containing the two duplicate numbers in any order.

Example: If the list should contain [0, 1, 2, 3] but instead contains [0, 1, 1, 2, 3, 3], then numbers 1 and 3 are the sneaky duplicates!

Input & Output

example_1.py — Basic Case

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

› Output: [1, 0]

💡 Note: The original array should be [0, 1] but numbers 1 and 0 each appear twice, making them the sneaky duplicates.

example_2.py — Sequential Duplicates

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

› Output: [3, 2]

💡 Note: The original array should be [0, 1, 2, 3] but numbers 3 and 2 appear twice each.

example_3.py — Minimum Case

$ Input: nums = [7, 1, 5, 4, 3, 4, 6, 0, 9, 5, 8, 2]

› Output: [4, 5]

💡 Note: In this larger array, numbers 4 and 5 are the two that appear twice among the range 0-9.

Visualization

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Understanding the Visualization

Setup Investigation

Prepare your checklist (hash set) and evidence bag (result array)

Interview Suspects

Go through each number in the lineup one by one

Cross-Reference

Check if each number is already on your seen list

Catch Duplicates

If already seen, it's a sneaky duplicate - add to evidence!

Update Records

If new, add to your seen list for future reference

Key Takeaway

🎯 Key Insight: Using a hash set allows us to detect duplicates in real-time with O(1) lookup operations, achieving optimal O(n) time complexity compared to the O(n²) brute force approach.

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array, with O(1) hash set operations

✓ Linear Growth

Space Complexity

O(n)

Hash set can store up to n-2 unique elements in worst case

⚡ Linearithmic Space

Constraints

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

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses a hash set to track seen numbers in a single pass. When we encounter a number already in the set, we've found a duplicate. This achieves O(n) time complexity with O(n) space complexity, making it much more efficient than the O(n²) brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Compare each element with every other element to find duplicates
One-Pass Hash Set (Optimal)	O(n)	O(n)	Use a hash set to track seen numbers and detect duplicates instantly

Brute Force (Nested Loops) — Algorithm Steps

Iterate through each element at index i
For each element, iterate through all elements at indices j > i
If nums[i] == nums[j], we found a duplicate
Add the duplicate to result array
Continue until we find both duplicates

Visualization

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Step-by-Step Walkthrough

Initialize

Start with the first element and compare it with all subsequent elements

Compare Pairs

Check each pair (i,j) where i < j for equality

Find Match

When nums[i] == nums[j], we found a duplicate

Continue

Repeat for all elements until both duplicates are found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int* getSneakyNumbers(int* nums, int numsSize, int* returnSize) {
    int* result = (int*)malloc(2 * sizeof(int));
    int count = 0;
    
    for (int i = 0; i < numsSize && count < 2; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] == nums[j]) {
                result[count++] = nums[i];
                break;
            }
        }
    }
    
    *returnSize = count;
    return result;
}

int main() {
    int nums[] = {0, 1, 1, 0}; // Example input
    int returnSize;
    int* result = getSneakyNumbers(nums, 4, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(", ");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of the n elements, we potentially check n-1 other elements

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a constant amount of extra space for the result array

✓ Linear Space

Constraints

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

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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