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

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            "title": "Minimum Operations to Reduce X to Zero",
            "titleSlug": "minimum-operations-to-reduce-x-to-zero",
            "content": "<p>You are given an integer array <code>nums</code> and an integer <code>x</code>. In one operation, you can either remove the leftmost or the rightmost element from the array <code>nums</code> and subtract its value from <code>x</code>. Note that this <strong>modifies</strong> the array for future operations.</p>\n\n<p>Return <em>the <strong>minimum number</strong> of operations to reduce </em><code>x</code> <em>to <strong>exactly</strong></em> <code>0</code> <em>if it is possible</em><em>, otherwise, return </em><code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,1,4,2,3], x = 5\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The optimal solution is to remove the last two elements to reduce x to zero.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,6,7,8,9], x = 4\n<strong>Output:</strong> -1\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,2,20,1,1,3], x = 10\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</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>4</sup></code></li>\n\t<li><code>1 &lt;= x &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "将 x 减到 0 的最小操作数",
            "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",
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