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"categoryTitle": "Algorithms",
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"title": "Maximum Product After K Increments",
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"content": "<p>You are given an array of non-negative integers <code>nums</code> and an integer <code>k</code>. In one operation, you may choose <strong>any</strong> element from <code>nums</code> and <strong>increment</strong> it by <code>1</code>.</p>\n\n<p>Return<em> the <strong>maximum</strong> <strong>product</strong> of </em><code>nums</code><em> after <strong>at most</strong> </em><code>k</code><em> operations. </em>Since the answer may be very large, return it <b>modulo</b> <code>10<sup>9</sup> + 7</code>. Note that you should maximize the product before taking the modulo.&nbsp;</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,4], k = 5\n<strong>Output:</strong> 20\n<strong>Explanation:</strong> Increment the first number 5 times.\nNow nums = [5, 4], with a product of 5 * 4 = 20.\nIt can be shown that 20 is maximum product possible, so we return 20.\nNote that there may be other ways to increment nums to have the maximum product.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [6,3,3,2], k = 2\n<strong>Output:</strong> 216\n<strong>Explanation:</strong> Increment the second number 1 time and increment the fourth number 1 time.\nNow nums = [6, 4, 3, 3], with a product of 6 * 4 * 3 * 3 = 216.\nIt can be shown that 216 is maximum product possible, so we return 216.\nNote that there may be other ways to increment nums to have the maximum product.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
"translatedTitle": "K 次增加后的最大乘积",
"translatedContent": "<p>给你一个非负整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;。每次操作,你可以选择&nbsp;<code>nums</code>&nbsp;中 <strong>任一</strong>&nbsp;元素并将它 <strong>增加</strong>&nbsp;<code>1</code>&nbsp;。</p>\n\n<p>请你返回 <strong>至多</strong>&nbsp;<code>k</code>&nbsp;次操作后,能得到的<em>&nbsp;</em><code>nums</code>的&nbsp;<strong>最大乘积</strong>&nbsp;。由于答案可能很大,请你将答案对&nbsp;<code>10<sup>9</sup> + 7</code>&nbsp;取余后返回。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><b>输入:</b>nums = [0,4], k = 5\n<b>输出:</b>20\n<b>解释:</b>将第一个数增加 5 次。\n得到 nums = [5, 4] ,乘积为 5 * 4 = 20 。\n可以证明 20 是能得到的最大乘积,所以我们返回 20 。\n存在其他增加 nums 的方法,也能得到最大乘积。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><b>输入:</b>nums = [6,3,3,2], k = 2\n<b>输出:</b>216\n<b>解释:</b>将第二个数增加 1 次,将第四个数增加 1 次。\n得到 nums = [6, 4, 3, 3] ,乘积为 6 * 4 * 3 * 3 = 216 。\n可以证明 216 是能得到的最大乘积,所以我们返回 216 。\n存在其他增加 nums 的方法,也能得到最大乘积。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
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"If you can increment only once, which number should you increment?",
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