N-Repeated Element in Size 2N Array

Problem Statement

In the given problem, we are presented with an integer array nums which has very specific characteristics. The array's length is twice that of a number n (i.e., 2 * n). The array nums is composed of exactly n + 1 unique elements. Among these unique elements, one specific element appears exactly n times. The task is to identify and return this repeating element.

Examples

Example 1

Input:

nums = [1,2,3,3]

Output:

3

Example 2

Input:

nums = [2,1,2,5,3,2]

Output:

2

Example 3

Input:

nums = [5,1,5,2,5,3,5,4]

Output:

5

Constraints

• 2 <= n <= 5000
• nums.length == 2 * n
• 0 <= nums[i] <= 104
• nums contains n + 1 unique elements and one of them is repeated exactly n times.

Approach and Intuition

The problem is about locating the element that repeats n times in the array of size 2 * n, which ensures that every other element is unique apart from the one that repeats. Let’s break down the strategy to tackle this:

1. Use a Hash Map:

   • Create a dictionary to store the frequency of each element as we iterate through the array.
   • For every element in the array, update its count in the dictionary.
   • Check if the count of the current element has reached n. If yes, this is the element we are looking for.

2. Observations and Optimization:

   • Considering only one element repeats n times and the length of the array is exactly 2 * n, this repeating element must appear in the first n + 1 elements. This insight can potentially reduce the number of elements we need to check.

3. Using the Constraints:

   • Given that 2 <= n <= 5000, the numerical operations remain computationally feasible within this range for methods involving linear traversal or hashing.
   • The constraint nums.length == 2 * n simplifies checking processes as we inherently know the structure regarding the distribution of element frequencies.

Examples Analysis

• In Example 1 (nums = [1,2,3,3]), iterating through the array, when we reach the second occurrence of 3, the count becomes 2. Since the array’s length is 4 (n = 2), the repeating element 3 is found eligible as it appears n times.

• Regarding Example 2 (nums = [2,1,2,5,3,2]), as we proceed through the array, the element 2 reaches the count of 3, which matches n, given the array size is 6 (hence, n = 3).

• In Example 3 (nums = [5,1,5,2,5,3,5,4]), the number 5 shows up multiple times and eventually hits the count 4, which corresponds to n for an array size 8.

The repeated integrations in these examples directly conform to the logic of either brute force through hashmap counting or an optimized approach based on deduced observations from the size constraints.

Solutions

class Solution {
    public int findRepeatedElement(int[] elements) {
        for (int distance = 1; distance <= 3; distance++)
            for (int index = 0; index < elements.length - distance; index++)
                if (elements[index] == elements[index + distance])
                    return elements[index];

        throw null;
    }
}