leetcode-problemset/leetcode-cn/problem (English)/半径为 k 的子数组平均值(English) [k-radius-subarray-averages].html

<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of <code>n</code> integers, and an integer <code>k</code>.</p>

<p>The <strong>k-radius average</strong> for a subarray of <code>nums</code> <strong>centered</strong> at some index <code>i</code> with the <strong>radius</strong> <code>k</code> is the average of <strong>all</strong> elements in <code>nums</code> between the indices <code>i - k</code> and <code>i + k</code> (<strong>inclusive</strong>). If there are less than <code>k</code> elements before <strong>or</strong> after the index <code>i</code>, then the <strong>k-radius average</strong> is <code>-1</code>.</p>

<p>Build and return <em>an array </em><code>avgs</code><em> of length </em><code>n</code><em> where </em><code>avgs[i]</code><em> is the <strong>k-radius average</strong> for the subarray centered at index </em><code>i</code>.</p>

<p>The <strong>average</strong> of <code>x</code> elements is the sum of the <code>x</code> elements divided by <code>x</code>, using <strong>integer division</strong>. The integer division truncates toward zero, which means losing its fractional part.</p>

<ul>
	<li>For example, the average of four elements <code>2</code>, <code>3</code>, <code>1</code>, and <code>5</code> is <code>(2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75</code>, which truncates to <code>2</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/11/07/eg1.png" style="width: 343px; height: 119px;" />
<pre>
<strong>Input:</strong> nums = [7,4,3,9,1,8,5,2,6], k = 3
<strong>Output:</strong> [-1,-1,-1,5,4,4,-1,-1,-1]
<strong>Explanation:</strong>
- avg[0], avg[1], and avg[2] are -1 because there are less than k elements <strong>before</strong> each index.
- The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.
  Using <strong>integer division</strong>, avg[3] = 37 / 7 = 5.
- For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.
- For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.
- avg[6], avg[7], and avg[8] are -1 because there are less than k elements <strong>after</strong> each index.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [100000], k = 0
<strong>Output:</strong> [100000]
<strong>Explanation:</strong>
- The sum of the subarray centered at index 0 with radius 0 is: 100000.
  avg[0] = 100000 / 1 = 100000.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [8], k = 100000
<strong>Output:</strong> [-1]
<strong>Explanation:</strong> 
- avg[0] is -1 because there are less than k elements before and after index 0.
</pre>

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

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