3289. The Two Sneaky Numbers of Digit ville

[mah-shamim]

Topics: Array, Hash Table, Math, Weekly Contest 415

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

Hint:

1. To solve the problem without the extra space, we need to think about how many times each number occurs in relation to the index.

[mah-shamim]

We need to find the two numbers that appear twice in an array that should contain each number from 0 to n-1 exactly once.

Let me analyze the approaches:

Approach 1: Using a Frequency Array

• Create an array to count occurrences of each number
• Find which numbers appear twice
• Time Complexity: O(n), Space Complexity: O(n)

Approach 2: Using Sum and Mathematical Properties

• Calculate expected sum vs actual sum
• Calculate expected sum of squares vs actual sum of squares
• Solve the system of equations to find the two numbers
• Time Complexity: O(n), Space Complexity: O(1)

Approach 3: Using Sorting

• Sort the array and find duplicates
• Time Complexity: O(n log n), Space Complexity: O(1) or O(n)

Let's implement this solution in PHP: 3289. The Two Sneaky Numbers of Digit ville

<?php
/**
 * @param Integer[] $nums
 * @return Integer[]
 */
function getSneakyNumbers($nums) {
    $n = count($nums) - 2;
    $freq = array_fill(0, $n, 0);
    $result = [];

    // Count frequency of each number
    foreach ($nums as $num) {
        $freq[$num]++;
    }

    // Find numbers that appear twice
    for ($i = 0; $i < $n; $i++) {
        if ($freq[$i] == 2) {
            $result[] = $i;
        }
    }

    return $result;
}

// Test cases
// Example 1
$nums = [0, 1, 1, 0];
print_r(getSneakyNumbers($nums)); // [0, 1]

// Example 2
$nums = [0, 3, 2, 1, 3, 2];
print_r(getSneakyNumbers($nums)); // [2, 3]

// Example 3
$nums = [7, 1, 5, 4, 3, 4, 6, 0, 9, 5, 8, 2];
print_r(getSneakyNumbers($nums)); // [4, 5]
?>

Explanation:

1. First, I calculate n by subtracting 2 from the length of nums (since the array has 2 extra elements)
2. I create a frequency array $freq initialized with zeros, with size n
3. I iterate through $nums and count occurrences of each number
4. I then iterate through the frequency array and find all numbers that appear exactly twice
5. These numbers are added to the result array and returned

Example walkthrough with nums = [0,1,1,0]:

• n = 4 - 2 = 2
• Frequency array starts as [0, 0]
• After counting: [2, 2] (0 appears twice, 1 appears twice)
• Both 0 and 1 are added to result: [0, 1]

This solution is efficient with O(n) time complexity and O(n) space complexity, which is optimal for the given constraints (n ≤ 100).

[topugit]

Thanks for solving the problem of "The Two Sneaky Numbers of Digitville"

[mah-shamim]

Thanks