# Solving Leetcode Interviews in Seconds with AI: N-Repeated Element in Size 2N Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "961" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given an integer array nums with the following properties:  nums.length == 2 * n. nums contains n + 1 unique elements. Exactly one element of nums is repeated n times.  Return the element that is repeated n times.   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.  

	# Explanation
	Here's a solution to the problem:

*   **High-Level Approach:**
    *   Iterate through the array.
    *   Use a hash map (dictionary in Python) to count the occurrences of each number.
    *   Return the number whose count equals `n`.

*   **Complexity Analysis:**
    *   Runtime Complexity: O(n), where n is the length of the input array `nums`.
    *   Storage Complexity: O(n), where n is the length of the input array `nums`, in the worst case (all elements are unique except for the repeated one).

	
	# Code
	```python
	def repeatedNTimes(nums: list[int]) -> int:
    """
    You are given an integer array nums with the following properties:
    nums.length == 2 * n.
    nums contains n + 1 unique elements.
    Exactly one element of nums is repeated n times.

    Return the element that is repeated n times.

    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.
    """
    counts = {}
    n = len(nums) // 2
    for num in nums:
        if num in counts:
            counts[num] += 1
            if counts[num] == n:
                return num
        else:
            counts[num] = 1
    return -1  # Should not reach here, but added for safety.
	```