# Solving Leetcode Interviews in Seconds with AI: Check If All 1's Are at Least Length K Places Away


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1437" 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 binary array nums and an integer k, return true if all 1's are at least k places away from each other, otherwise return false.   Example 1:   Input: nums = [1,0,0,0,1,0,0,1], k = 2 Output: true Explanation: Each of the 1s are at least 2 places away from each other.  Example 2:   Input: nums = [1,0,0,1,0,1], k = 2 Output: false Explanation: The second 1 and third 1 are only one apart from each other.    Constraints:  1 <= nums.length <= 105 0 <= k <= nums.length nums[i] is 0 or 1  

	# Explanation
	*   **Iterate and Track:** Iterate through the `nums` array. Keep track of the index of the last encountered '1'.
*   **Distance Check:** For each '1' encountered, check if the distance from the previous '1' is at least `k`.
*   **Early Exit:** If the distance is ever less than `k`, immediately return `False`. Otherwise, if the loop completes, return `True`.

*   **Time Complexity:** O(n), where n is the length of the nums array. **Space Complexity:** O(1).

	
	# Code
	```python
	def kLengthApart(nums, k):
    """
    Checks if all 1's in a binary array are at least k places away from each other.

    Args:
        nums: A list of integers representing the binary array.
        k: An integer representing the minimum distance between 1's.

    Returns:
        True if all 1's are at least k places away from each other, False otherwise.
    """

    last_one = float('-inf')  # Initialize with negative infinity to handle the first '1'
    for i, num in enumerate(nums):
        if num == 1:
            if i - last_one - 1 < k:
                return False
            last_one = i
    return True
	```