leetcode-problemset/leetcode-cn/originData/slowest-key.json

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        "question": {
            "questionId": "1751",
            "questionFrontendId": "1629",
            "categoryTitle": "Algorithms",
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            "title": "Slowest Key",
            "titleSlug": "slowest-key",
            "content": "<p>A newly designed keypad was tested, where a tester pressed a sequence of <code>n</code> keys, one at a time.</p>\n\n<p>You are given a string <code>keysPressed</code> of length <code>n</code>, where <code>keysPressed[i]</code> was the <code>i<sup>th</sup></code> key pressed in the testing sequence, and a sorted list <code>releaseTimes</code>, where <code>releaseTimes[i]</code> was the time the <code>i<sup>th</sup></code> key was released. Both arrays are <strong>0-indexed</strong>. The <code>0<sup>th</sup></code> key was pressed at the time <code>0</code>,&nbsp;and every subsequent key was pressed at the <strong>exact</strong> time the previous key was released.</p>\n\n<p>The tester wants to know the key of the keypress that had the <strong>longest duration</strong>. The <code>i<sup>th</sup></code><sup> </sup>keypress had a <strong>duration</strong> of <code>releaseTimes[i] - releaseTimes[i - 1]</code>, and the <code>0<sup>th</sup></code> keypress had a duration of <code>releaseTimes[0]</code>.</p>\n\n<p>Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key <strong>may not</strong> have had the same <strong>duration</strong>.</p>\n\n<p><em>Return the key of the keypress that had the <strong>longest duration</strong>. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.</em></p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> releaseTimes = [9,29,49,50], keysPressed = &quot;cbcd&quot;\n<strong>Output:</strong> &quot;c&quot;\n<strong>Explanation:</strong> The keypresses were as follows:\nKeypress for &#39;c&#39; had a duration of 9 (pressed at time 0 and released at time 9).\nKeypress for &#39;b&#39; had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).\nKeypress for &#39;c&#39; had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).\nKeypress for &#39;d&#39; had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).\nThe longest of these was the keypress for &#39;b&#39; and the second keypress for &#39;c&#39;, both with duration 20.\n&#39;c&#39; is lexicographically larger than &#39;b&#39;, so the answer is &#39;c&#39;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> releaseTimes = [12,23,36,46,62], keysPressed = &quot;spuda&quot;\n<strong>Output:</strong> &quot;a&quot;\n<strong>Explanation:</strong> The keypresses were as follows:\nKeypress for &#39;s&#39; had a duration of 12.\nKeypress for &#39;p&#39; had a duration of 23 - 12 = 11.\nKeypress for &#39;u&#39; had a duration of 36 - 23 = 13.\nKeypress for &#39;d&#39; had a duration of 46 - 36 = 10.\nKeypress for &#39;a&#39; had a duration of 62 - 46 = 16.\nThe longest of these was the keypress for &#39;a&#39; with duration 16.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>releaseTimes.length == n</code></li>\n\t<li><code>keysPressed.length == n</code></li>\n\t<li><code>2 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= releaseTimes[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>releaseTimes[i] &lt; releaseTimes[i+1]</code></li>\n\t<li><code>keysPressed</code> contains only lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "按键持续时间最长的键",
            "translatedContent": "<p>LeetCode 设计了一款新式键盘，正在测试其可用性。测试人员将会点击一系列键（总计 <code>n</code> 个），每次一个。</p>\n\n<p>给你一个长度为 <code>n</code> 的字符串 <code>keysPressed</code> ，其中 <code>keysPressed[i]</code> 表示测试序列中第 <code>i</code> 个被按下的键。<code>releaseTimes</code> 是一个升序排列的列表，其中 <code>releaseTimes[i]</code> 表示松开第 <code>i</code> 个键的时间。字符串和数组的 <strong>下标都从 0 开始</strong> 。第 <code>0</code> 个键在时间为 <code>0</code> 时被按下，接下来每个键都 <strong>恰好</strong> 在前一个键松开时被按下。</p>\n\n<p>测试人员想要找出按键 <strong>持续时间最长</strong> 的键。第 <code>i</code><sup> </sup>次按键的持续时间为 <code>releaseTimes[i] - releaseTimes[i - 1]</code> ，第 <code>0</code> 次按键的持续时间为 <code>releaseTimes[0]</code> 。</p>\n\n<p>注意，测试期间，同一个键可以在不同时刻被多次按下，而每次的持续时间都可能不同。</p>\n\n<p>请返回单次按键 <strong>持续时间最长</strong> 的键，如果有多个这样的键，则返回 <strong>按字母顺序排列最大</strong> 的那个键。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>releaseTimes = [9,29,49,50], keysPressed = \"cbcd\"\n<strong>输出：</strong>\"c\"\n<strong>解释：</strong>按键顺序和持续时间如下：\n按下 'c' ，持续时间 9（时间 0 按下，时间 9 松开）\n按下 'b' ，持续时间 29 - 9 = 20（松开上一个键的时间 9 按下，时间 29 松开）\n按下 'c' ，持续时间 49 - 29 = 20（松开上一个键的时间 29 按下，时间 49 松开）\n按下 'd' ，持续时间 50 - 49 = 1（松开上一个键的时间 49 按下，时间 50 松开）\n按键持续时间最长的键是 'b' 和 'c'（第二次按下时），持续时间都是 20\n'c' 按字母顺序排列比 'b' 大，所以答案是 'c'\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>releaseTimes = [12,23,36,46,62], keysPressed = \"spuda\"\n<strong>输出：</strong>\"a\"\n<strong>解释：</strong>按键顺序和持续时间如下：\n按下 's' ，持续时间 12\n按下 'p' ，持续时间 23 - 12 = 11\n按下 'u' ，持续时间 36 - 23 = 13\n按下 'd' ，持续时间 46 - 36 = 10\n按下 'a' ，持续时间 62 - 46 = 16\n按键持续时间最长的键是 'a' ，持续时间 16</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>releaseTimes.length == n</code></li>\n\t<li><code>keysPressed.length == n</code></li>\n\t<li><code>2 &lt;= n &lt;= 1000</code></li>\n\t<li><code>1 &lt;= releaseTimes[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>releaseTimes[i] &lt; releaseTimes[i+1]</code></li>\n\t<li><code>keysPressed</code> 仅由小写英文字母组成</li>\n</ul>\n",
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                "Get for each press its key and amount of time taken.",
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