05. Minimize the Heights II

The problem can be found at the following link: Question Link

My Approach

Simply greddy approach, I start by sorting the array. Then, for each element, I consider two possibilities:

• Getting max after adding k to the current element and subtracting k from the maximum element.

• Getting min after subtracting k from the current element and adding k to the minimum element. I update the minimum difference (out) by comparing it with the difference obtained from the above two possibilities.

Time and Auxiliary Space Complexity

• Time Complexity: O(nlog n) for sorting arr array.

• Auxiliary Space Complexity: O(1), as no extra space is used.

Code (C++)

class Solution {
public:
    int getMinDiff(int arr[], int n, int k) {
        sort(arr, arr + n);
        int out = arr[n - 1] - arr[0];

        for (int i = 0; i < n - 1; ++i) {
            if (arr[i + 1] - k >= 0) {
                int nax = max(arr[i] + k, arr[n - 1] - k);
                int nin = min(arr[i + 1] - k, arr[0] + k);
                out = min(out, nax - nin);
            }
        }
        return out;
    }
};

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