# Solving Leetcode Interviews in Seconds with AI: Find All Lonely Numbers in the Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2150" 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. A number x is lonely when it appears only once, and no adjacent numbers (i.e. x + 1 and x - 1) appear in the array. Return all lonely numbers in nums. You may return the answer in any order.   Example 1:  Input: nums = [10,6,5,8] Output: [10,8] Explanation:  - 10 is a lonely number since it appears exactly once and 9 and 11 does not appear in nums. - 8 is a lonely number since it appears exactly once and 7 and 9 does not appear in nums. - 5 is not a lonely number since 6 appears in nums and vice versa. Hence, the lonely numbers in nums are [10, 8]. Note that [8, 10] may also be returned.  Example 2:  Input: nums = [1,3,5,3] Output: [1,5] Explanation:  - 1 is a lonely number since it appears exactly once and 0 and 2 does not appear in nums. - 5 is a lonely number since it appears exactly once and 4 and 6 does not appear in nums. - 3 is not a lonely number since it appears twice. Hence, the lonely numbers in nums are [1, 5]. Note that [5, 1] may also be returned.    Constraints:  1 <= nums.length <= 105 0 <= nums[i] <= 106  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Counting:** Use a dictionary (hash map) to store the frequency of each number in the input array.
*   **Filtering:** Iterate through the frequency map. If a number appears only once, check if its neighbors (number + 1 and number - 1) exist in the frequency map. If they don't, the number is lonely.
*   **Collecting:** Add all identified lonely numbers to a result list and return.

*   **Runtime Complexity:** O(n), where n is the length of the input array.
*   **Storage Complexity:** O(n) in the worst case, to store the frequency map.

	
	# Code
	```python
	def findLonely(nums):
    """
    Finds and returns all lonely numbers in the input array.

    Args:
      nums: A list of integers.

    Returns:
      A list of lonely numbers.
    """

    freq = {}
    for num in nums:
        freq[num] = freq.get(num, 0) + 1

    lonely_numbers = []
    for num, count in freq.items():
        if count == 1:
            if (num + 1) not in freq and (num - 1) not in freq:
                lonely_numbers.append(num)

    return lonely_numbers
	```