update

leetcode-cn/originData/maximum-distance-between-a-pair-of-values.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你两个 <strong>非递增</strong> 的整数数组 <code>nums1</code>​​​​​​ 和 <code>nums2</code>​​​​​​ ，数组下标均 <strong>从 0 开始</strong> 计数。</p>\n\n<p>下标对 <code>(i, j)</code> 中 <code>0 &lt;= i &lt; nums1.length</code> 且 <code>0 &lt;= j &lt; nums2.length</code> 。如果该下标对同时满足 <code>i &lt;= j</code> 且 <code>nums1[i] &lt;= nums2[j]</code> ，则称之为 <strong>有效</strong> 下标对，该下标对的 <strong>距离</strong> 为 <code>j - i</code>​​ 。​​</p>\n\n<p>返回所有 <strong>有效</strong> 下标对<em> </em><code>(i, j)</code><em> </em>中的 <strong>最大距离</strong> 。如果不存在有效下标对，返回 <code>0</code> 。</p>\n\n<p>一个数组 <code>arr</code> ，如果每个 <code>1 &lt;= i &lt; arr.length</code> 均有 <code>arr[i-1] &gt;= arr[i]</code> 成立，那么该数组是一个 <strong>非递增</strong> 数组。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]\n<strong>输出：</strong>2\n<strong>解释：</strong>有效下标对是 (0,0), (2,2), (2,3), (2,4), (3,3), (3,4) 和 (4,4) 。\n最大距离是 2 ，对应下标对 (2,4) 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [2,2,2], nums2 = [10,10,1]\n<strong>输出：</strong>1\n<strong>解释：</strong>有效下标对是 (0,0), (0,1) 和 (1,1) 。\n最大距离是 1 ，对应下标对 (0,1) 。</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]\n<strong>输出：</strong>2\n<strong>解释：</strong>有效下标对是 (2,2), (2,3), (2,4), (3,3) 和 (3,4) 。\n最大距离是 2 ，对应下标对 (2,4) 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums2.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums1[i], nums2[j] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>nums1</code> 和 <code>nums2</code> 都是 <strong>非递增</strong> 数组</li>\n</ul>\n",
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             "hints": [
                 "Since both arrays are sorted in a non-increasing way this means that for each value in the first array. We can find the farthest value smaller than it using binary search.",
                 "There is another solution using a two pointers approach since the first array is non-increasing the farthest j such that nums2[j] ≥ nums1[i] is at least as far as the farthest j such that nums2[j] ≥ nums1[i-1]"