<p>Given an binary array <code>nums</code> and an integer <code>k</code>, return <code>true</code><em> if all </em><code>1</code><em>&#39;s are at least </em><code>k</code><em> places away from each other, otherwise return </em><code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png" style="width: 428px; height: 181px;" />
<pre>
<strong>Input:</strong> nums = [1,0,0,0,1,0,0,1], k = 2
<strong>Output:</strong> true
<strong>Explanation:</strong> Each of the 1s are at least 2 places away from each other.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png" style="width: 320px; height: 173px;" />
<pre>
<strong>Input:</strong> nums = [1,0,0,1,0,1], k = 2
<strong>Output:</strong> false
<strong>Explanation:</strong> The second 1 and third 1 are only one apart from each other.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
	<li><code>nums[i]</code> is <code>0</code> or <code>1</code></li>
</ul>

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        i = -k-1
        for j in range(len(nums)):
            if nums[j] == 1:
                if j-i <= k:
                    return False
                i = j
        return True
436