C++ Recovering an Array from a Given 2n Array Using Sorting and Hash Maps
Intuition\n\nThe problem requires recovering an array of n integers from a given array of 2n integers such that each element in the original array is the midpoi
# Intuition\n\nThe problem requires recovering an array of n integers from a given array of 2n integers such that each element in the original array is the midpoint of two elements in the given array. We need to determine a value k such that when adding and subtracting k from each element in the recovered array, the result matches the given array. Using sorting and hash maps, we can efficiently track and verify pairs to reconstruct the original array.\n\n# Approach\nSort the Array: Begin by sorting the given array to simplify finding pairs.\nIterate Over Potential k: For each possible k (determined by examining pairs of elements in the sorted array), check if it can be used to form the required array.\nValidation of k: Use a hash map to track the frequency of each number and verify if we can consistently find pairs (x-k, x+k) in the sorted array. If we can form such pairs for the entire array, we have found the correct k.\nReturn the Result: If a valid k is found, return the recovered array; otherwise, continue until all possibilities are exhausted.\n\n# Complexity\n- Time complexity:\n$$O(nlogn)$$ for sorting the array and $$O(n)$$ for the hash map operations within the loop. Thus, the overall time complexity is $$O(nlogn)$$.\n\n- Space complexity:\n$$O(n)$$ for storing the hash map and temporary vectors.\n\n# Code\n```\nclass Solution {\npublic:\n    vector<int> recoverArray(vector<int>& nums) {\n        sort(nums.begin(), nums.end());\n        int n = nums.size() / 2;\n\n        for (int start = 1; start < nums.size(); ++start) {\n            int potential_k = (nums[start] - nums[0]) / 2;\n\n            if (potential_k <= 0 || (nums[start] - nums[0]) % 2 != 0) {\n                continue;\n            }\n\n            unordered_map<int, int> count;\n            for (int num : nums) {\n                count[num]++;\n            }\n\n            bool valid = true;\n            vector<int> tempArray;\n\n            for (int num : nums) {\n                if (count[num] == 0) continue;\n                if (count[num + 2 * potential_k] == 0) {\n                    valid = false;\n                    break;\n                }\n                tempArray.push_back(num + potential_k);\n                count[num]--;\n                count[num + 2 * potential_k]--;\n            }\n\n            if (valid) {\n                return tempArray;\n            }\n        }\n\n        return {};\n    }\n};\n\n```
['C++']