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            "title": "Slowest Key",
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            "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",
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                    "code": "class Solution {\npublic:\n    char slowestKey(vector<int>& releaseTimes, string keysPressed) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def slowestKey(self, releaseTimes, keysPressed):\n        \"\"\"\n        :type releaseTimes: List[int]\n        :type keysPressed: str\n        :rtype: str\n        \"\"\"\n        ",
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                    "code": "/**\n * @param {number[]} releaseTimes\n * @param {string} keysPressed\n * @return {character}\n */\nvar slowestKey = function(releaseTimes, keysPressed) {\n    \n};",
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                    "code": "# @param {Integer[]} release_times\n# @param {String} keys_pressed\n# @return {Character}\ndef slowest_key(release_times, keys_pressed)\n    \nend",
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                    "code": "impl Solution {\n    pub fn slowest_key(release_times: Vec<i32>, keys_pressed: String) -> char {\n        \n    }\n}",
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                "Get for each press its key and amount of time taken.",
                "Iterate on the presses, maintaining the answer so far.",
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