# Solving Leetcode Interviews in Seconds with AI: Majority Element


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "169" 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
	> Given an array nums of size n, return the majority element. The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.   Example 1: Input: nums = [3,2,3] Output: 3 Example 2: Input: nums = [2,2,1,1,1,2,2] Output: 2    Constraints:  n == nums.length 1 <= n <= 5 * 104 -109 <= nums[i] <= 109    Follow-up: Could you solve the problem in linear time and in O(1) space?

	# Explanation
	*   **Boyer-Moore Voting Algorithm:** This algorithm cancels out elements that are not the majority element. We maintain a `candidate` and a `count`. If the current element is the same as the `candidate`, we increment the `count`; otherwise, we decrement it. If the `count` becomes zero, we choose a new `candidate`.
*   **Linear Scan:** The algorithm iterates through the array only once, making it efficient.
*   **Guaranteed Majority:** Since the problem states that a majority element always exists, the final `candidate` will be the majority element.

*   **Time Complexity:** O(n), **Space Complexity:** O(1)

	
	# Code
	```python
	def majorityElement(nums):
    """
    Finds the majority element in an array using Boyer-Moore Voting Algorithm.

    Args:
        nums: A list of integers.

    Returns:
        The majority element in the array.
    """
    candidate = 0
    count = 0

    for num in nums:
        if count == 0:
            candidate = num
        if num == candidate:
            count += 1
        else:
            count -= 1

    return candidate
	```