leetcode-problemset/leetcode/originData/maximum-score-from-performing-multiplication-operations.json

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            "title": "Maximum Score from Performing Multiplication Operations",
            "titleSlug": "maximum-score-from-performing-multiplication-operations",
            "content": "<p>You are given two integer arrays <code>nums</code> and <code>multipliers</code><strong> </strong>of size <code>n</code> and <code>m</code> respectively, where <code>n &gt;= m</code>. The arrays are <strong>1-indexed</strong>.</p>\n\n<p>You begin with a score of <code>0</code>. You want to perform <strong>exactly</strong> <code>m</code> operations. On the <code>i<sup>th</sup></code> operation <strong>(1-indexed)</strong>, you will:</p>\n\n<ul>\n\t<li>Choose one integer <code>x</code> from <strong>either the start or the end </strong>of the array <code>nums</code>.</li>\n\t<li>Add <code>multipliers[i] * x</code> to your score.</li>\n\t<li>Remove <code>x</code> from the array <code>nums</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>maximum</strong> score after performing </em><code>m</code> <em>operations.</em></p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3], multipliers = [3,2,1]\n<strong>Output:</strong> 14\n<strong>Explanation:</strong>&nbsp;An optimal solution is as follows:\n- Choose from the end, [1,2,<strong><u>3</u></strong>], adding 3 * 3 = 9 to the score.\n- Choose from the end, [1,<strong><u>2</u></strong>], adding 2 * 2 = 4 to the score.\n- Choose from the end, [<strong><u>1</u></strong>], adding 1 * 1 = 1 to the score.\nThe total score is 9 + 4 + 1 = 14.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]\n<strong>Output:</strong> 102\n<strong>Explanation: </strong>An optimal solution is as follows:\n- Choose from the start, [<u><strong>-5</strong></u>,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.\n- Choose from the start, [<strong><u>-3</u></strong>,-3,-2,7,1], adding -3 * -5 = 15 to the score.\n- Choose from the start, [<strong><u>-3</u></strong>,-2,7,1], adding -3 * 3 = -9 to the score.\n- Choose from the end, [-2,7,<strong><u>1</u></strong>], adding 1 * 4 = 4 to the score.\n- Choose from the end, [-2,<strong><u>7</u></strong>], adding 7 * 6 = 42 to the score. \nThe total score is 50 + 15 - 9 + 4 + 42 = 102.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>m == multipliers.length</code></li>\n\t<li><code>1 &lt;= m &lt;= 10<sup>3</sup></code></li>\n\t<li><code>m &lt;= n &lt;= 10<sup>5</sup></code><code> </code></li>\n\t<li><code>-1000 &lt;= nums[i], multipliers[i] &lt;= 1000</code></li>\n</ul>\n",
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                    "code": "class Solution(object):\n    def maximumScore(self, nums, multipliers):\n        \"\"\"\n        :type nums: List[int]\n        :type multipliers: List[int]\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "/**\n * @param {number[]} nums\n * @param {number[]} multipliers\n * @return {number}\n */\nvar maximumScore = function(nums, multipliers) {\n    \n};",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer[]} multipliers\n# @return {Integer}\ndef maximum_score(nums, multipliers)\n    \nend",
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                    "code": "func maximumScore(nums []int, multipliers []int) int {\n    \n}",
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                    "code": "object Solution {\n    def maximumScore(nums: Array[Int], multipliers: Array[Int]): Int = {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn maximum_score(nums: Vec<i32>, multipliers: Vec<i32>) -> i32 {\n        \n    }\n}",
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                    "code": "-spec maximum_score(Nums :: [integer()], Multipliers :: [integer()]) -> integer().\nmaximum_score(Nums, Multipliers) ->\n  .",
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            "hints": [
                "At first glance, the solution seems to be greedy, but if you try to greedily take the largest value from the beginning or the end, this will not be optimal.",
                "You should try all scenarios but this will be costy.",
                "Memoizing the pre-visited states while trying all the possible scenarios will reduce the complexity, and hence dp is a perfect choice here."
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