mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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189 lines
20 KiB
JSON
189 lines
20 KiB
JSON
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"title": "Earliest Possible Day of Full Bloom",
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"content": "<p>You have <code>n</code> flower seeds. Every seed must be planted first before it can begin to grow, then bloom. Planting a seed takes time and so does the growth of a seed. You are given two <strong>0-indexed</strong> integer arrays <code>plantTime</code> and <code>growTime</code>, of length <code>n</code> each:</p>\n\n<ul>\n\t<li><code>plantTime[i]</code> is the number of <strong>full days</strong> it takes you to <strong>plant</strong> the <code>i<sup>th</sup></code> seed. Every day, you can work on planting exactly one seed. You <strong>do not</strong> have to work on planting the same seed on consecutive days, but the planting of a seed is not complete <strong>until</strong> you have worked <code>plantTime[i]</code> days on planting it in total.</li>\n\t<li><code>growTime[i]</code> is the number of <strong>full days</strong> it takes the <code>i<sup>th</sup></code> seed to grow after being completely planted. <strong>After</strong> the last day of its growth, the flower <strong>blooms</strong> and stays bloomed forever.</li>\n</ul>\n\n<p>From the beginning of day <code>0</code>, you can plant the seeds in <strong>any</strong> order.</p>\n\n<p>Return <em>the <strong>earliest</strong> possible day where <strong>all</strong> seeds are blooming</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/21/1.png\" style=\"width: 453px; height: 149px;\" />\n<pre>\n<strong>Input:</strong> plantTime = [1,4,3], growTime = [2,3,1]\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms.\nOne optimal way is:\nOn day 0, plant the 0<sup>th</sup> seed. The seed grows for 2 full days and blooms on day 3.\nOn days 1, 2, 3, and 4, plant the 1<sup>st</sup> seed. The seed grows for 3 full days and blooms on day 8.\nOn days 5, 6, and 7, plant the 2<sup>nd</sup> seed. The seed grows for 1 full day and blooms on day 9.\nThus, on day 9, all the seeds are blooming.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/21/2.png\" style=\"width: 454px; height: 184px;\" />\n<pre>\n<strong>Input:</strong> plantTime = [1,2,3,2], growTime = [2,1,2,1]\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms.\nOne optimal way is:\nOn day 1, plant the 0<sup>th</sup> seed. The seed grows for 2 full days and blooms on day 4.\nOn days 0 and 3, plant the 1<sup>st</sup> seed. The seed grows for 1 full day and blooms on day 5.\nOn days 2, 4, and 5, plant the 2<sup>nd</sup> seed. The seed grows for 2 full days and blooms on day 8.\nOn days 6 and 7, plant the 3<sup>rd</sup> seed. The seed grows for 1 full day and blooms on day 9.\nThus, on day 9, all the seeds are blooming.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> plantTime = [1], growTime = [1]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> On day 0, plant the 0<sup>th</sup> seed. The seed grows for 1 full day and blooms on day 2.\nThus, on day 2, all the seeds are blooming.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == plantTime.length == growTime.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= plantTime[i], growTime[i] <= 10<sup>4</sup></code></li>\n</ul>\n",
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"List the planting like the diagram above shows, where a row represents the timeline of a seed. A row i is above another row j if the last day planting seed i is ahead of the last day for seed j. Does it have any advantage to spend some days to plant seed j before completely planting seed i?",
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"No. It does not help seed j but could potentially delay the completion of seed i, resulting in a worse final answer. Remaining focused is a part of the optimal solution.",
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"Sort the seeds by their growTime in descending order. Can you prove why this strategy is the other part of the optimal solution? Note the bloom time of a seed is the sum of plantTime of all seeds preceding this seed plus the growTime of this seed.",
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"There is no way to improve this strategy. The seed to bloom last dominates the final answer. Exchanging the planting of this seed with another seed with either a larger or smaller growTime will result in a potentially worse answer."
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