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leetcode-cn/originData/make-sum-divisible-by-p.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个正整数数组&nbsp;<code>nums</code>，请你移除 <strong>最短</strong>&nbsp;子数组（可以为 <strong>空</strong>），使得剩余元素的 <strong>和</strong>&nbsp;能被 <code>p</code>&nbsp;整除。 <strong>不允许</strong>&nbsp;将整个数组都移除。</p>\n\n<p>请你返回你需要移除的最短子数组的长度，如果无法满足题目要求，返回 <code>-1</code>&nbsp;。</p>\n\n<p><strong>子数组</strong>&nbsp;定义为原数组中连续的一组元素。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>nums = [3,1,4,2], p = 6\n<strong>输出：</strong>1\n<strong>解释：</strong>nums 中元素和为 10，不能被 p 整除。我们可以移除子数组 [4] ，剩余元素的和为 6 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>nums = [6,3,5,2], p = 9\n<strong>输出：</strong>2\n<strong>解释：</strong>我们无法移除任何一个元素使得和被 9 整除，最优方案是移除子数组 [5,2] ，剩余元素为 [6,3]，和为 9 。\n</pre>\n\n<p><strong>示例&nbsp;3：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1,2,3], p = 3\n<strong>输出：</strong>0\n<strong>解释：</strong>和恰好为 6 ，已经能被 3 整除了。所以我们不需要移除任何元素。\n</pre>\n\n<p><strong>示例&nbsp; 4：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1,2,3], p = 7\n<strong>输出：</strong>-1\n<strong>解释：</strong>没有任何方案使得移除子数组后剩余元素的和被 7 整除。\n</pre>\n\n<p><strong>示例 5：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1000000000,1000000000,1000000000], p = 3\n<strong>输出：</strong>0\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 &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= p &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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             "hints": [
                 "Use prefix sums to calculate the subarray sums.",
                 "Suppose you know the remainder for the sum of the entire array. How does removing a subarray affect that remainder? What remainder does the subarray need to have in order to make the rest of the array sum up to be divisible by k?",