<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>

const int offset = 5e4+5;
int N = -1e5;

class Solution {
public:
    vector<int> sortArray(vector<int>& a) {
        int n = a.size();
        
        for(int i = 0; i < n; ++i){
            N = max(N,a[i]);
        }

        vector<int>cnt(N+1+offset);

        for(int i = 0; i < n; ++i){
            cnt[a[i] + offset]++;
        }

        vector<int>ans;

        for(int i = 0; i < N+1+offset; ++i){
            for(int j = 0; j < cnt[i]; ++j){
                ans.push_back(i-offset);
            }
        }

        return ans;
    }
};
80986