Python -  easy and intuitive approach using sorting and slide window. O(nlogn)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n\n1. Search for values already in nums:\nFor each value n in nums, determ
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n\n1. Search for values already in nums:\nFor each value n in nums, determine the range of values within k units of n. This range includes values in nums that can potentially be converted to n. First, calculate how many times n appears (ori_count) and count the other values within this range. Use numOperations to maximize the frequency of n by converting as many of these other values as possible to n. Track the highest frequency achieved.\n\n2. Search for values not in nums:\nFor target values not currently in nums, each frequency count must be achieved by converting values through numOperations. For each possible new target value within a range of 2 * k units from values in nums, calculate the maximum frequency achievable using up to numOperations operations.\n\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def maxFrequency(self, nums: List[int], k: int, numOperations: int) -> int:\n        nums.sort()\n        counter = Counter(nums)\n        ans = max(counter.values())\n\n        left = 0\n        right = 0\n        #Search for values already in nums:\n        for idx, n in enumerate(nums):\n            while(nums[left] < n-k):\n                left += 1\n            while(right + 1 < len(nums) and nums[right + 1] <= n + k):\n                right += 1\n\n            ori_count = counter[n]\n            other = right - left + 1 - ori_count\n            ans = max(ans, ori_count + min(numOperations, other))\n\n\n        right = 0\n        #Search for values not in nums\n        for left, left_n in enumerate(nums):\n            while(right + 1 < len(nums) and nums[right + 1] <= left_n + 2 * k):\n                right += 1\n            ans = max(ans, min(right-left+1, numOperations))\n            \n        return ans\n```
['Python3']