leetcode-problemset/leetcode-cn/problem (English)/超过阈值的最少操作数 I(English) [minimum-operations-to-exceed-threshold-value-i].html

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

<p>In one operation, you can remove one occurrence of the smallest element of <code>nums</code>.</p>

<p>Return <em>the <strong>minimum</strong> number of operations needed so that all elements of the array are greater than or equal to</em> <code>k</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,11,10,1,3], k = 10
<strong>Output:</strong> 3
<strong>Explanation:</strong> After one operation, nums becomes equal to [2, 11, 10, 3].
After two operations, nums becomes equal to [11, 10, 3].
After three operations, nums becomes equal to [11, 10].
At this stage, all the elements of nums are greater than or equal to 10 so we can stop.
It can be shown that 3 is the minimum number of operations needed so that all elements of the array are greater than or equal to 10.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,2,4,9], k = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> All elements of the array are greater than or equal to 1 so we do not need to apply any operations on nums.</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,2,4,9], k = 9
<strong>Output:</strong> 4
<strong>Explanation:</strong> only a single element of nums is greater than or equal to 9 so we need to apply the operations 4 times on nums.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
	<li>The input is generated such that there is at least one index <code>i</code> such that <code>nums[i] &gt;= k</code>.</li>
</ul>