Easy to Understand || C++ Code || Explaination
Intuition\n Describe your first thoughts on how to solve this problem. \nIntervals Count\n\n# Approach\n Describe your approach to solving the problem. \nEach e
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntervals Count\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nEach element can be expanded by adding or subtracting k, forming an interval `[x\u2212k,x+k]`. \nTo find the maximum number of overlapping intervals, represent these ranges as start and end points and sort them.\nAdditionally add the element with indentifier so that it comes up in between performing the sorting.\n\nTraverse the sorted array while maintaining a count of active intervals to identify the maximum overlap. When aiming to transform other elements in the array to match a specific number \uD835\uDC65, consider its existing frequency. Subtract this frequency from the count since these instances don\u2019t need transformation.\n\n# Complexity\n- Time complexity: O(n * log(n))\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> \n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int maxFrequency(vector<int>& nums, int k, int numOperations) {\n        int n = nums.size();\n        vector <pair<int, int>> possibles;\n        map <int ,int> freq;\n        for (int i = 0; i < n; i++) {\n            possibles.push_back({nums[i] - k, 0});\n            possibles.push_back({nums[i], 1});\n            possibles.push_back({nums[i] + k, 2});\n            freq[nums[i]] += 1;\n        }\n        int ans = 0;\n        sort(possibles.begin(), possibles.end());\n        int cnt = 0;\n        for (int i = 0; i < possibles.size(); i++) {\n            if (possibles[i].second == 0) {\n                cnt += 1;\n            }\n            else if (possibles[i].second == 2){\n                cnt -= 1;\n            }\n            if(possibles[i].second == 1) {\n                ans = max(ans, min(cnt - freq[possibles[i].first], numOperations) + freq[possibles[i].first]);\n            }\n            ans = max(ans, min(cnt, numOperations));\n            // cout << i << " " <<  possibles[i].first << " " << possibles[i].second <<" "<< ans << "\\n";\n        }\n        return ans;\n    }\n};\n```
['C++']