update

leetcode-cn/originData/minimum-operations-to-reduce-x-to-zero.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>x</code> 。每一次操作时，你应当移除数组 <code>nums</code> 最左边或最右边的元素，然后从 <code>x</code> 中减去该元素的值。请注意，需要 <strong>修改</strong> 数组以供接下来的操作使用。</p>\n\n<p>如果可以将 <code>x</code> <strong>恰好</strong> 减到 <code>0</code> ，返回<strong> 最小操作数 </strong>；否则，返回 <code>-1</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,1,4,2,3], x = 5\n<strong>输出：</strong>2\n<strong>解释：</strong>最佳解决方案是移除后两个元素，将 x 减到 0 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [5,6,7,8,9], x = 4\n<strong>输出：</strong>-1\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,2,20,1,1,3], x = 10\n<strong>输出：</strong>5\n<strong>解释：</strong>最佳解决方案是移除后三个元素和前两个元素（总共 5 次操作），将 x 减到 0 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= 10<sup>4</sup></code></li>\n\t<li><code>1 <= x <= 10<sup>9</sup></code></li>\n</ul>\n",
             "isPaidOnly": false,
             "difficulty": "Medium",
-            "likes": 90,
+            "likes": 97,
             "dislikes": 0,
             "isLiked": null,
             "similarQuestions": "[]",
@@ -161,7 +161,7 @@
                     "__typename": "CodeSnippetNode"
                 }
             ],
-            "stats": "{\"totalAccepted\": \"11.4K\", \"totalSubmission\": \"36.6K\", \"totalAcceptedRaw\": 11437, \"totalSubmissionRaw\": 36622, \"acRate\": \"31.2%\"}",
+            "stats": "{\"totalAccepted\": \"11.9K\", \"totalSubmission\": \"37.9K\", \"totalAcceptedRaw\": 11854, \"totalSubmissionRaw\": 37923, \"acRate\": \"31.3%\"}",
             "hints": [
                 "Think in reverse; instead of finding the minimum prefix + suffix, find the maximum subarray.",
                 "Finding the maximum subarray is standard and can be done greedily."