mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
47 lines
2.0 KiB
HTML
47 lines
2.0 KiB
HTML
<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>value</code>.</p>
|
|
|
|
<p>In one operation, you can add or subtract <code>value</code> from any element of <code>nums</code>.</p>
|
|
|
|
<ul>
|
|
<li>For example, if <code>nums = [1,2,3]</code> and <code>value = 2</code>, you can choose to subtract <code>value</code> from <code>nums[0]</code> to make <code>nums = [-1,2,3]</code>.</li>
|
|
</ul>
|
|
|
|
<p>The MEX (minimum excluded) of an array is the smallest missing <strong>non-negative</strong> integer in it.</p>
|
|
|
|
<ul>
|
|
<li>For example, the MEX of <code>[-1,2,3]</code> is <code>0</code> while the MEX of <code>[1,0,3]</code> is <code>2</code>.</li>
|
|
</ul>
|
|
|
|
<p>Return <em>the maximum MEX of </em><code>nums</code><em> after applying the mentioned operation <strong>any number of times</strong></em>.</p>
|
|
|
|
<p> </p>
|
|
<p><strong class="example">Example 1:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> nums = [1,-10,7,13,6,8], value = 5
|
|
<strong>Output:</strong> 4
|
|
<strong>Explanation:</strong> One can achieve this result by applying the following operations:
|
|
- Add value to nums[1] twice to make nums = [1,<strong><u>0</u></strong>,7,13,6,8]
|
|
- Subtract value from nums[2] once to make nums = [1,0,<strong><u>2</u></strong>,13,6,8]
|
|
- Subtract value from nums[3] twice to make nums = [1,0,2,<strong><u>3</u></strong>,6,8]
|
|
The MEX of nums is 4. It can be shown that 4 is the maximum MEX we can achieve.
|
|
</pre>
|
|
|
|
<p><strong class="example">Example 2:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> nums = [1,-10,7,13,6,8], value = 7
|
|
<strong>Output:</strong> 2
|
|
<strong>Explanation:</strong> One can achieve this result by applying the following operation:
|
|
- subtract value from nums[2] once to make nums = [1,-10,<u><strong>0</strong></u>,13,6,8]
|
|
The MEX of nums is 2. It can be shown that 2 is the maximum MEX we can achieve.
|
|
</pre>
|
|
|
|
<p> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>1 <= nums.length, value <= 10<sup>5</sup></code></li>
|
|
<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
|
|
</ul>
|