leetcode-problemset/leetcode/originData/apply-operations-to-maximize-frequency-score.json

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            "title": "Apply Operations to Maximize Frequency Score",
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            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> and an integer <code>k</code>.</p>\n\n<p>You can perform the following operation on the array <strong>at most</strong> <code>k</code> times:</p>\n\n<ul>\n\t<li>Choose any index <code>i</code> from the array and <strong>increase</strong> or <strong>decrease</strong> <code>nums[i]</code> by <code>1</code>.</li>\n</ul>\n\n<p>The score of the final array is the <strong>frequency</strong> of the most frequent element in the array.</p>\n\n<p>Return <em>the <strong>maximum</strong> score you can achieve</em>.</p>\n\n<p>The frequency of an element is the number of occurences of that element in the array.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,6,4], k = 3\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We can do the following operations on the array:\n- Choose i = 0, and increase the value of nums[0] by 1. The resulting array is [2,2,6,4].\n- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,3].\n- Choose i = 3, and decrease the value of nums[3] by 1. The resulting array is [2,2,6,2].\nThe element 2 is the most frequent in the final array so our score is 3.\nIt can be shown that we cannot achieve a better score.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,4,4,2,4], k = 0\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We cannot apply any operations so our score will be the frequency of the most frequent element in the original array, which is 3.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= k &lt;= 10<sup>14</sup></code></li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Solution {\npublic:\n    int maxFrequencyScore(vector<int>& nums, long long k) {\n        \n    }\n};",
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                    "lang": "Python",
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                    "code": "class Solution(object):\n    def maxFrequencyScore(self, nums, k):\n        \"\"\"\n        :type nums: List[int]\n        :type k: int\n        :rtype: int\n        \"\"\"\n        ",
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                    "lang": "C",
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                    "code": "int maxFrequencyScore(int* nums, int numsSize, long long k) {\n    \n}",
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                    "code": "public class Solution {\n    public int MaxFrequencyScore(int[] nums, long k) {\n        \n    }\n}",
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                    "code": "/**\n * @param {number[]} nums\n * @param {number} k\n * @return {number}\n */\nvar maxFrequencyScore = function(nums, k) {\n    \n};",
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                    "code": "function maxFrequencyScore(nums: number[], k: number): number {\n    \n};",
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                    "code": "class Solution {\n\n    /**\n     * @param Integer[] $nums\n     * @param Integer $k\n     * @return Integer\n     */\n    function maxFrequencyScore($nums, $k) {\n        \n    }\n}",
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                    "code": "class Solution {\n    func maxFrequencyScore(_ nums: [Int], _ k: Int) -> Int {\n        \n    }\n}",
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                    "code": "class Solution {\n    fun maxFrequencyScore(nums: IntArray, k: Long): Int {\n        \n    }\n}",
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                    "code": "func maxFrequencyScore(nums []int, k int64) int {\n    \n}",
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                    "code": "# @param {Integer[]} nums\n# @param {Integer} k\n# @return {Integer}\ndef max_frequency_score(nums, k)\n    \nend",
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                    "code": "impl Solution {\n    pub fn max_frequency_score(nums: Vec<i32>, k: i64) -> i32 {\n        \n    }\n}",
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            "hints": [
                "If you sort the original array, it is optimal to apply the operations on one subarray such that all the elements of that subarray become equal.",
                "You can use binary search to find the longest subarray where we can make the elements equal in at most <code>k</code> operations."
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