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leetcode-problemset/leetcode-cn/problem (English)/质数减法运算(English) [prime-subtraction-operation].html
2023-03-26 12:40:26 +08:00

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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>
<p>You can perform the following operation as many times as you want:</p>
<ul>
<li>Pick an index <code>i</code> that you haven&rsquo;t picked before, and pick a prime <code>p</code> <strong>strictly less than</strong> <code>nums[i]</code>, then subtract <code>p</code> from <code>nums[i]</code>.</li>
</ul>
<p>Return <em>true if you can make <code>nums</code> a strictly increasing array using the above operation and false otherwise.</em></p>
<p>A <strong>strictly increasing array</strong> is an array whose each element is strictly greater than its preceding element.</p>
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [4,9,6,10]
<strong>Output:</strong> true
<strong>Explanation:</strong> In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0], so that nums becomes [1,9,6,10].
In the second operation: i = 1, p = 7, subtract 7 from nums[1], so nums becomes equal to [1,2,6,10].
After the second operation, nums is sorted in strictly increasing order, so the answer is true.</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [6,8,11,12]
<strong>Output:</strong> true
<strong>Explanation: </strong>Initially nums is sorted in strictly increasing order, so we don&#39;t need to make any operations.</pre>
<p><strong class="example">Example 3:</strong></p>
<pre>
<strong>Input:</strong> nums = [5,8,3]
<strong>Output:</strong> false
<strong>Explanation:</strong> It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
<li><code><font face="monospace">nums.length == n</font></code></li>
</ul>