count-days-without-meetings

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Algorithms

You are given a positive integer <code>days</code> representing the total number of days an employee is available for work (starting from day 1). You are also given a 2D array <code>meetings</code> of size <code>n</code> where, <code>meetings[i] = [start_i, end_i]</code> represents the starting and ending days of meeting <code>i</code> (inclusive).

Return the count of days when the employee is available for work but no meetings are scheduled.

Note: The meetings may overlap.

&nbsp;
Example 1:

<div class="example-block">
Input: days = 10, meetings = [[5,7],[1,3],[9,10]]

Output: 2

Explanation:

There is no meeting scheduled on the 4th and 8th days.
</div>

Example 2:

<div class="example-block">
Input: days = 5, meetings = [[2,4],[1,3]]

Output: 1

Explanation:

There is no meeting scheduled on the 5th day.
</div>

Example 3:

<div class="example-block">
Input: days = 6, meetings = [[1,6]]

Output: 0

Explanation:

Meetings are scheduled for all working days.
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= days &lt;= 109</code></li>
	<li><code>1 &lt;= meetings.length &lt;= 105</code></li>
	<li><code>meetings[i].length == 2</code></li>
	<li><code>1 &lt;= meetings[i][0] &lt;= meetings[i][1] &lt;= days</code></li>
</ul>

countDays

Count Days Without Meetings

Merge Intervals

merge-intervals

合并区间

Array

Sorting

给你一个正整数 <code>days</code>，表示员工可工作的总天数（从第 1 天开始）。另给你一个二维数组 <code>meetings</code>，长度为 <code>n</code>，其中 <code>meetings[i] = [start_i, end_i]</code> 表示第 <code>i</code> 次会议的开始和结束天数（包含首尾）。

返回员工可工作且没有安排会议的天数。

注意：会议时间可能会有重叠。

&nbsp;

示例 1：

<div class="example-block">
输入：days = 10, meetings = [[5,7],[1,3],[9,10]]

输出：2

解释：

第 4 天和第 8 天没有安排会议。
</div>

示例 2：

<div class="example-block">
输入：days = 5, meetings = [[2,4],[1,3]]

输出：1

解释：

第 5 天没有安排会议。
</div>

示例 3：

<div class="example-block">
输入：days = 6, meetings = [[1,6]]

输出：0

解释：

所有工作日都安排了会议。
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= days &lt;= 109</code></li>
	<li><code>1 &lt;= meetings.length &lt;= 105</code></li>
	<li><code>meetings[i].length == 2</code></li>
	<li><code>1 &lt;= meetings[i][0] &lt;= meetings[i][1] &lt;= days</code></li>
</ul>