<p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class countingSort {
private:
    vector<int> vec;
    vector<int> bucket = vector<int>(1e5, 0);
    int shift;

public:
    countingSort (vector<int> nums) {
        vec = vector<int>(nums.begin(), nums.end());
        shift = -ranges::min(vec);
    }

    void sort() {
        int left = 0, right = vec.size() - 1;
        sort(left, right);
    }

    void sort(int left, int right) {
        
        for (int idx = left ; idx <= right ; ++idx ) {
            bucket[vec[idx] + shift]++;
        }

        int idx = left;
        for (int i = 0 ; i < 1e5 ; ++i ) {
            while (bucket[i]-- > 0) {
                vec[idx++] = i - shift; 
            }
        }
    }

    vector<int> getData() const {
        return vec;
    }
};

class Solution {
public:
    vector<int> sortArray(vector<int>& nums) {
        
        countingSort cs(nums);
        cs.sort();

        return cs.getData();
    }
};
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