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LC - E - 912. Sort an Array

Solutions in C++

Solution 1

Merge sort.

Complexity Analysis:

• T = O(n log n)

• S = O(n)

Solution:

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class Solution {
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public:
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    vector<int> sortArray(vector<int>& nums) {
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        // base case        
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        if (nums.size() <= 1)
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            return nums;
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        int mid = nums.size() / 2;
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        // split the original array into halves
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        vector<int> left(nums.begin(), nums.begin() + mid);
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        vector<int> right(nums.begin() + mid, nums.end());
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        sortArray(left);
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        sortArray(right);
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        merge(nums, left, right);
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        return nums;
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    }
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private:
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    void merge(vector<int> &arr, const vector<int> &left, const vector<int> &right)
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    {
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        int i = 0;  // index for the original array
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        int l = 0;  // index for the left subarray
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        int r = 0;  // index for the right subarray
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        // memrge elements of the two subarrays
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        while (l < left.size() && r < right.size())
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        {
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            arr[i++] = (left[l] <= right[r]) ? left[l++] : right[r++];
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        }
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        // copy remaining elements (only one subarray may have leftovers)
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        while (l < left.size())
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        {
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            arr[i++] = left[l++];
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        }
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        while (r < right.size())
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        {
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            arr[i++] = right[r++];
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        }
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    }
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};