leetcode-problemset/leetcode/originData/minimum-time-to-remove-all-cars-containing-illegal-goods.json

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            "title": "Minimum Time to Remove All Cars Containing Illegal Goods",
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            "content": "<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> which represents a sequence of train cars. <code>s[i] = &#39;0&#39;</code> denotes that the <code>i<sup>th</sup></code> car does <strong>not</strong> contain illegal goods and <code>s[i] = &#39;1&#39;</code> denotes that the <code>i<sup>th</sup></code> car does contain illegal goods.</p>\n\n<p>As the train conductor, you would like to get rid of all the cars containing illegal goods. You can do any of the following three operations <strong>any</strong> number of times:</p>\n\n<ol>\n\t<li>Remove a train car from the <strong>left</strong> end (i.e., remove <code>s[0]</code>) which takes 1 unit of time.</li>\n\t<li>Remove a train car from the <strong>right</strong> end (i.e., remove <code>s[s.length - 1]</code>) which takes 1 unit of time.</li>\n\t<li>Remove a train car from <strong>anywhere</strong> in the sequence which takes 2 units of time.</li>\n</ol>\n\n<p>Return <em>the <strong>minimum</strong> time to remove all the cars containing illegal goods</em>.</p>\n\n<p>Note that an empty sequence of cars is considered to have no cars containing illegal goods.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;<strong><u>11</u></strong>00<strong><u>1</u></strong>0<strong><u>1</u></strong>&quot;\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> \nOne way to remove all the cars containing illegal goods from the sequence is to\n- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.\n- remove a car from the right end. Time taken is 1.\n- remove the car containing illegal goods found in the middle. Time taken is 2.\nThis obtains a total time of 2 + 1 + 2 = 5. \n\nAn alternative way is to\n- remove a car from the left end 2 times. Time taken is 2 * 1 = 2.\n- remove a car from the right end 3 times. Time taken is 3 * 1 = 3.\nThis also obtains a total time of 2 + 3 = 5.\n\n5 is the minimum time taken to remove all the cars containing illegal goods. \nThere are no other ways to remove them with less time.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;00<strong><u>1</u></strong>0&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\nOne way to remove all the cars containing illegal goods from the sequence is to\n- remove a car from the left end 3 times. Time taken is 3 * 1 = 3.\nThis obtains a total time of 3.\n\nAnother way to remove all the cars containing illegal goods from the sequence is to\n- remove the car containing illegal goods found in the middle. Time taken is 2.\nThis obtains a total time of 2.\n\nAnother way to remove all the cars containing illegal goods from the sequence is to \n- remove a car from the right end 2 times. Time taken is 2 * 1 = 2. \nThis obtains a total time of 2.\n\n2 is the minimum time taken to remove all the cars containing illegal goods. \nThere are no other ways to remove them with less time.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>\n</ul>\n",
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            "hints": [
                "Build an array withoutFirst where withoutFirst[i] stores the minimum time to remove all the cars containing illegal goods from the ‘suffix’ of the sequence starting from the ith car without using any type 1 operations.",
                "Next, build an array onlyFirst where onlyFirst[i] stores the minimum time to remove all the cars containing illegal goods from the ‘prefix’ of the sequence ending on the ith car using only type 1 operations.",
                "Finally, we can compare the best way to split the operations amongst these two types by finding the minimum time across all onlyFirst[i] + withoutFirst[i + 1]."
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