The frequency of an element is the number of times it occurs in an array.
\n\nYou are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1.
Return the maximum possible frequency of an element after performing at most k operations.
\n
Example 1:
\n\n\nInput: nums = [1,2,4], k = 5\nOutput: 3\nExplanation: Increment the first element three times and the second element two times to make nums = [4,4,4].\n4 has a frequency of 3.\n\n
Example 2:
\n\n\nInput: nums = [1,4,8,13], k = 5\nOutput: 2\nExplanation: There are multiple optimal solutions:\n- Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.\n- Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.\n- Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.\n\n\n
Example 3:
\n\n\nInput: nums = [3,9,6], k = 2\nOutput: 1\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1051 <= k <= 105元素的 频数 是该元素在一个数组中出现的次数。
\n\n给你一个整数数组 nums 和一个整数 k 。在一步操作中，你可以选择 nums 的一个下标，并将该下标对应元素的值增加 1 。
执行最多 k 次操作后，返回数组中最高频元素的 最大可能频数 。
\n\n
示例 1：
\n\n\n输入：nums = [1,2,4], k = 5\n输出：3\n解释：对第一个元素执行 3 次递增操作，对第二个元素执 2 次递增操作，此时 nums = [4,4,4] 。\n4 是数组中最高频元素，频数是 3 。\n\n
示例 2：
\n\n\n输入：nums = [1,4,8,13], k = 5\n输出：2\n解释：存在多种最优解决方案：\n- 对第一个元素执行 3 次递增操作，此时 nums = [4,4,8,13] 。4 是数组中最高频元素，频数是 2 。\n- 对第二个元素执行 4 次递增操作，此时 nums = [1,8,8,13] 。8 是数组中最高频元素，频数是 2 。\n- 对第三个元素执行 5 次递增操作，此时 nums = [1,4,13,13] 。13 是数组中最高频元素，频数是 2 。\n\n\n
示例 3：
\n\n\n输入：nums = [3,9,6], k = 2\n输出：1\n\n\n
\n\n
提示：
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