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, 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

    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