check-if-all-1s-are-at-least-length-k-places-away

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Algorithms

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

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/15/sample_1_1791.png" style="width: 428px; height: 181px;" />
<pre>
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.
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2020/04/15/sample_2_1791.png" style="width: 320px; height: 173px;" />
<pre>
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.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</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>

kLengthApart

Check If All 1's Are at Least Length K Places Away

Task Scheduler II

task-scheduler-ii

任务调度器 II

Array

给你一个由若干 <code>0</code> 和 <code>1</code> 组成的数组 <code>nums</code> 以及整数 <code>k</code>。如果所有 <code>1</code> 都至少相隔 <code>k</code> 个元素，则返回 true&nbsp;；否则，返回 <code>false</code> 。

&nbsp;

示例 1：

<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/05/03/sample_1_1791.png" style="width: 214px;" />

<pre>
输入：nums = [1,0,0,0,1,0,0,1], k = 2
输出：true
解释：每个 1 都至少相隔 2 个元素。</pre>

示例 2：

<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/05/03/sample_2_1791.png" style="height: 86px; width: 160px;" />

<pre>
输入：nums = [1,0,0,1,0,1], k = 2
输出：false
解释：第二个 1 和第三个 1 之间只隔了 1 个元素。</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 105</code></li>
	<li><code>0 &lt;= k &lt;= nums.length</code></li>
	<li><code>nums[i]</code> 的值为 <code>0</code> 或 <code>1</code></li>
</ul>