leetcode-problemset/leetcode-cn/originData/maximum-energy-boost-from-two-drinks.json

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        "question": {
            "questionId": "3525",
            "questionFrontendId": "3259",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2883077,
            "title": "Maximum Energy Boost From Two Drinks",
            "titleSlug": "maximum-energy-boost-from-two-drinks",
            "content": "<p>You are given two integer arrays <code>energyDrinkA</code> and <code>energyDrinkB</code> of the same length <code>n</code> by a futuristic sports scientist. These arrays represent the energy boosts per hour provided by two different energy drinks, A and B, respectively.</p>\n\n<p>You want to <em>maximize</em> your total energy boost by drinking one energy drink <em>per hour</em>. However, if you want to switch from consuming one energy drink to the other, you need to wait for <em>one hour</em> to cleanse your system (meaning you won&#39;t get any energy boost in that hour).</p>\n\n<p>Return the <strong>maximum</strong> total energy boost you can gain in the next <code>n</code> hours.</p>\n\n<p><strong>Note</strong> that you can start consuming <em>either</em> of the two energy drinks.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> energyDrinkA<span class=\"example-io\"> = [1,3,1], </span>energyDrinkB<span class=\"example-io\"> = [3,1,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>To gain an energy boost of 5, drink only the energy drink A (or only B).</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> energyDrinkA<span class=\"example-io\"> = [4,1,1], </span>energyDrinkB<span class=\"example-io\"> = [1,1,3]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">7</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>To gain an energy boost of 7:</p>\n\n<ul>\n\t<li>Drink the energy drink A for the first hour.</li>\n\t<li>Switch to the energy drink B and we lose the energy boost of the second hour.</li>\n\t<li>Gain the energy boost of the drink B in the third hour.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == energyDrinkA.length == energyDrinkB.length</code></li>\n\t<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= energyDrinkA[i], energyDrinkB[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "超级饮料的最大强化能量",
            "translatedContent": "<p>来自未来的体育科学家给你两个整数数组 <code>energyDrinkA</code> 和 <code>energyDrinkB</code>，数组长度都等于 <code>n</code>。这两个数组分别代表 A、B 两种不同能量饮料每小时所能提供的强化能量。</p>\n\n<p>你需要每小时饮用一种能量饮料来 <strong>最大化 </strong>你的总强化能量。然而，如果从一种能量饮料切换到另一种，你需要等待一小时来梳理身体的能量体系（在那个小时里你将不会获得任何强化能量）。</p>\n\n<p>返回在接下来的 <code>n</code> 小时内你能获得的<strong> 最大 </strong>总强化能量。</p>\n\n<p><strong>注意 </strong>你可以选择从饮用任意一种能量饮料开始。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong>energyDrinkA<span class=\"example-io\"> = [1,3,1], </span>energyDrinkB<span class=\"example-io\"> = [3,1,1]</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">5</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>要想获得 5 点强化能量，需要选择只饮用能量饮料 A（或者只饮用 B）。</p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong>energyDrinkA<span class=\"example-io\"> = [4,1,1], </span>energyDrinkB<span class=\"example-io\"> = [1,1,3]</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">7</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>第一个小时饮用能量饮料 A。</li>\n\t<li>切换到能量饮料 B ，在第二个小时无法获得强化能量。</li>\n\t<li>第三个小时饮用能量饮料 B ，并获得强化能量。</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == energyDrinkA.length == energyDrinkB.length</code></li>\n\t<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= energyDrinkA[i], energyDrinkB[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    long long maxEnergyBoost(vector<int>& energyDrinkA, vector<int>& energyDrinkB) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def maxEnergyBoost(self, energyDrinkA, energyDrinkB):\n        \"\"\"\n        :type energyDrinkA: List[int]\n        :type energyDrinkB: List[int]\n        :rtype: int\n        \"\"\"",
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                    "code": "public class Solution {\n    public long MaxEnergyBoost(int[] energyDrinkA, int[] energyDrinkB) {\n\n    }\n}",
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                    "code": "/**\n * @param {number[]} energyDrinkA\n * @param {number[]} energyDrinkB\n * @return {number}\n */\nvar maxEnergyBoost = function(energyDrinkA, energyDrinkB) {\n\n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $energyDrinkA\n     * @param Integer[] $energyDrinkB\n     * @return Integer\n     */\n    function maxEnergyBoost($energyDrinkA, $energyDrinkB) {\n\n    }\n}",
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                    "code": "class Solution {\n    fun maxEnergyBoost(energyDrinkA: IntArray, energyDrinkB: IntArray): Long {\n\n    }\n}",
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                    "code": "class Solution {\n  int maxEnergyBoost(List<int> energyDrinkA, List<int> energyDrinkB) {\n    \n  }\n}",
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                    "code": "object Solution {\n    def maxEnergyBoost(energyDrinkA: Array[Int], energyDrinkB: Array[Int]): Long = {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn max_energy_boost(energy_drink_a: Vec<i32>, energy_drink_b: Vec<i32>) -> i64 {\n\n    }\n}",
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            "hints": [
                "Can we solve it using dynamic programming?",
                "Define <code>dpA[i]</code> as the maximum energy boost if we consider only the first <code>i + 1</code> hours such that in the last hour, we drink the energy drink A.",
                "Similarly define <code>dpB[i]</code>.",
                "<code>dpA[i] = max(dpA[i - 1], dpB[i - 2]) + energyDrinkA[i]</code>",
                "Similarly, fill <code>dpB</code>.",
                "The answer is <code>max(dpA[n - 1], dpB[n - 1])</code>."
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