leetcode-problemset/leetcode/originData/maximum-split-of-positive-even-integers.json

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            "title": "Maximum Split of Positive Even Integers",
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            "content": "<p>You are given an integer <code>finalSum</code>. Split it into a sum of a <strong>maximum</strong> number of <strong>unique</strong> positive even integers.</p>\n\n<ul>\n\t<li>For example, given <code>finalSum = 12</code>, the following splits are <strong>valid</strong> (unique positive even integers summing up to <code>finalSum</code>): <code>(12)</code>, <code>(2 + 10)</code>, <code>(2 + 4 + 6)</code>, and <code>(4 + 8)</code>. Among them, <code>(2 + 4 + 6)</code> contains the maximum number of integers. Note that <code>finalSum</code> cannot be split into <code>(2 + 2 + 4 + 4)</code> as all the numbers should be unique.</li>\n</ul>\n\n<p>Return <em>a list of integers that represent a valid split containing a <strong>maximum</strong> number of integers</em>. If no valid split exists for <code>finalSum</code>, return <em>an <strong>empty</strong> list</em>. You may return the integers in <strong>any</strong> order.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> finalSum = 12\n<strong>Output:</strong> [2,4,6]\n<strong>Explanation:</strong> The following are valid splits: <code>(12)</code>, <code>(2 + 10)</code>, <code>(2 + 4 + 6)</code>, and <code>(4 + 8)</code>.\n(2 + 4 + 6) has the maximum number of integers, which is 3. Thus, we return [2,4,6].\nNote that [2,6,4], [6,2,4], etc. are also accepted.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> finalSum = 7\n<strong>Output:</strong> []\n<strong>Explanation:</strong> There are no valid splits for the given finalSum.\nThus, we return an empty array.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> finalSum = 28\n<strong>Output:</strong> [6,8,2,12]\n<strong>Explanation:</strong> The following are valid splits: <code>(2 + 26)</code>, <code>(6 + 8 + 2 + 12)</code>, and <code>(4 + 24)</code>. \n<code>(6 + 8 + 2 + 12)</code> has the maximum number of integers, which is 4. Thus, we return [6,8,2,12].\nNote that [10,2,4,12], [6,2,4,16], etc. are also accepted.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= finalSum &lt;= 10<sup>10</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<long long> maximumEvenSplit(long long finalSum) {\n        \n    }\n};",
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                    "code": "class Solution {\n    public List<Long> maximumEvenSplit(long finalSum) {\n        \n    }\n}",
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                    "code": "class Solution(object):\n    def maximumEvenSplit(self, finalSum):\n        \"\"\"\n        :type finalSum: int\n        :rtype: List[int]\n        \"\"\"\n        ",
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                    "code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nlong long* maximumEvenSplit(long long finalSum, int* returnSize){\n\n}",
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                    "code": "public class Solution {\n    public IList<long> MaximumEvenSplit(long finalSum) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number} finalSum\n * @return {number[]}\n */\nvar maximumEvenSplit = function(finalSum) {\n    \n};",
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                    "code": "# @param {Integer} final_sum\n# @return {Integer[]}\ndef maximum_even_split(final_sum)\n    \nend",
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                    "code": "class Solution {\n    func maximumEvenSplit(_ finalSum: Int) -> [Int] {\n        \n    }\n}",
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                    "code": "impl Solution {\n    pub fn maximum_even_split(final_sum: i64) -> Vec<i64> {\n        \n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer $finalSum\n     * @return Integer[]\n     */\n    function maximumEvenSplit($finalSum) {\n        \n    }\n}",
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            "hints": [
                "First, check if finalSum is divisible by 2. If it isn’t, then we cannot split it into even integers.",
                "Let k be the number of elements in our split. As we want the maximum number of elements, we should try to use the first k - 1 even elements to grow our sum as slowly as possible.",
                "Thus, we find the maximum sum of the first k - 1 even elements which is less than finalSum.",
                "We then add the difference over to the kth element."
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