Simple sliding window approach
\n### Explanation\n\n1. Initial Setup and Sorting:\n   - First, the array nums is sorted in ascending order. This allows us to work with contiguous subarrays wh
\n### Explanation\n\n1. **Initial Setup and Sorting**:\n   - First, the array `nums` is sorted in ascending order. This allows us to work with contiguous subarrays where elements are close in value, making it easier to consider converting elements to a single target value within a specific range.\n   - A frequency map (like a hash map or dictionary) is used to store the count of each element in `nums`. This helps us calculate potential maximum frequencies for each target value without modifying the array directly.\n\n2. **First Pass: Using Existing Numbers as Reference Points**:\n   - In this first pass, each element in `nums` is considered as a potential target frequency value. We try to make as many elements in a contiguous subarray equal to this target by adjusting values within a given range.\n   - For each element `nums[mid]` (considered as a reference point), a sliding window is created with `left` and `right` pointers. This window expands to include elements close enough in value such that they can potentially be converted to `nums[mid]` by adding or subtracting up to `k`.\n   - We adjust `left` and `right` pointers to keep the range within `k` of `nums[mid]`, ensuring all elements in this window could be converted to `nums[mid]`.\n\n   - **Formula**: The formula `(total - count[nums[mid]])` represents the number of elements in the window that would need adjustment to match `nums[mid]`. We take the minimum of this value and `numOperations` (to avoid exceeding the number of allowed operations), and then add `count[nums[mid]]` (the current frequency of `nums[mid]`) to get the maximum achievable frequency with `nums[mid]` as the target.\n\n3. **Second Pass: Using a Non-Existent Number as Reference Point**:\n   - In this second pass, instead of an existing number, we consider a hypothetical target value (a midpoint) that doesn\u2019t necessarily exist in `nums`. This approach can potentially increase frequency by creating a target that might be more centrally located within a subarray.\n   - We calculate this midpoint as `(nums[left] + nums[right]) / 2` and adjust `left` so that the midpoint is within `k` distance from both `nums[left]` and `nums[right]`. This ensures all elements between `left` and `right` can potentially be modified to match this midpoint.\n   - This pass allows for flexibility in choosing a target that might maximize frequency without relying on existing elements.\n\n4. **Updating the Maximum Frequency (`ans`)**:\n   - After each pass, we update `ans` with the maximum frequency achieved. This ensures that we capture the highest possible frequency from both the first (using existing numbers) and the second pass (using a hypothetical midpoint).\n\n### Complexity Analysis\n\n- **Time Complexity**: Sorting the array takes \\(O(n log n)\\), and each pass over the array is \\(O(n)\\), so the overall time complexity is \\(O(n log n)\\).\n- **Space Complexity**: \\(O(n)\\) is required for the frequency map used to store element counts.\n\n---\n\n### Summary\n\nThis code employs a two-part strategy:\n   - **First Pass**: Uses existing elements in `nums` as reference points.\n   - **Second Pass**: Uses a non-existent midpoint as a hypothetical reference point.\n\nThis combination of strategies provides flexibility in maximizing the frequency of a target value with minimal modifications, leveraging both existing and hypothetical reference points.\n\n# Code\n```java []\nclass Solution {\n    public int maxFrequency(int[] nums, int k, int numOperations) {\n        int n = nums.length, ans = 0, left = 0, right = 0;\n        Arrays.sort(nums);\n\n        // HashMap to store the frequency of each number in the sorted array\n        HashMap<Integer, Integer> count = new HashMap<>();\n        for(int num : nums) count.put(num, count.getOrDefault(num, 0) + 1);\n\n        // First pass: choose an existing number as the "reference point"\n        for(int mid = 0; mid < n; mid++) {\n            // Move the left pointer to maintain that the difference between nums[mid] and nums[left] is <= k\n            while(nums[mid] - nums[left] > k) {\n                left++;\n            }\n\n            // Move the right pointer to maintain that the difference between nums[right] and nums[mid] is <= k\n            while(right < n - 1 && nums[right + 1] - nums[mid] <= k) {\n                right++;\n            }\n\n            int total = right - left + 1; // total elements in the current range\n            ans = Math.max(ans, Math.min(total - count.get(nums[mid]), numOperations) + count.get(nums[mid]));\n        }\n\n        // Second pass: choose a non-existent number as the "reference point"\n        left = 0;\n        for(right = 0; right < n; right++) {\n            int mid = (nums[left] + nums[right]) / 2;\n\n            // Move the left pointer until the mid-point falls within range [nums[left], nums[right]]\n            while(mid - nums[left] > k || nums[right] - mid > k) {\n                left++;\n                mid = (nums[left] + nums[right]) / 2;\n            }\n            ans = Math.max(ans, Math.min(right - left + 1, numOperations));\n        }\n\n        return ans;\n    }\n}\n```\n\n```C++ []\nclass Solution {\npublic:\n    int maxFrequency(vector<int>& nums, int k, int numOperations) {\n        int n = nums.size(), ans = 0, left = 0, right = 0;\n        sort(nums.begin(), nums.end());  // Sort the array for sliding window approach\n\n        // HashMap to store frequency of each number\n        unordered_map<int, int> count;\n        for (int num : nums) {\n            count[num]++;\n        }\n\n        // First pass: choose an existing number as the reference point\n        for (int mid = 0; mid < n; mid++) {\n            // Adjust left pointer to keep nums[mid] - nums[left] within `k`\n            while (nums[mid] - nums[left] > k) {\n                left++;\n            }\n\n            // Adjust right pointer to keep nums[right] - nums[mid] within `k`\n            while (right < n - 1 && nums[right + 1] - nums[mid] <= k) {\n                right++;\n            }\n\n            int total = right - left + 1;  // Calculate range size\n            // Update ans with maximum achievable frequency for nums[mid] as the target\n            ans = max(ans, min(total - count[nums[mid]], numOperations) + count[nums[mid]]);\n        }\n\n        // Second pass: choose a non-existent number as reference point\n        left = 0;\n        for (right = 0; right < n; right++) {\n            int mid = (nums[left] + nums[right]) / 2;  // Calculate hypothetical midpoint\n\n            // Adjust left pointer to ensure midpoint is within `k` range from both ends\n            while (mid - nums[left] > k || nums[right] - mid > k) {\n                left++;\n                mid = (nums[left] + nums[right]) / 2;\n            }\n\n            // Update ans with maximum achievable frequency with the hypothetical midpoint\n            ans = max(ans, min(right - left + 1, numOperations));\n        }\n\n        return ans;\n    }\n}\n```\n\n```Python []\nclass Solution(object):\n    def maxFrequency(self, nums, k, numOperations):\n        # Sort the array to use a sliding window approach\n        nums.sort()\n        n = len(nums)\n        ans = 0\n        left = 0\n        right = 0\n        \n        # First pass: choose existing numbers as reference points\n        from collections import Counter\n        count = Counter(nums)  # Dictionary to store frequency of each element\n\n        for mid in range(n):\n            # Move the left pointer to maintain the range within `k`\n            while nums[mid] - nums[left] > k:\n                left += 1\n\n            # Move the right pointer to include numbers within `k` range from `nums[mid]`\n            while right < n - 1 and nums[right + 1] - nums[mid] <= k:\n                right += 1\n\n            # Calculate total elements in the current range\n            total = right - left + 1\n            # Update ans with the maximum frequency achievable with current reference\n            ans = max(ans, min(total - count[nums[mid]], numOperations) + count[nums[mid]])\n\n        # Second pass: choose a non-existent midpoint as reference\n        left = 0\n        for right in range(n):\n            # Calculate midpoint between nums[left] and nums[right]\n            mid = (nums[left] + nums[right]) // 2\n            # Move left pointer to keep midpoint within `k` range of both ends\n            while mid - nums[left] > k or nums[right] - mid > k:\n                left += 1\n                mid = (nums[left] + nums[right]) // 2\n            \n            # Update ans with maximum frequency achievable with non-existent midpoint\n            ans = max(ans, min(right - left + 1, numOperations))\n        \n        return ans\n\n```
['Python', 'C++', 'Java']