<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 Solution {
public:
    vector<int> sortArray(vector<int>& nums) {
        const int base = 1e5;
        vector<vector<int>> cnt(10, vector<int>());
        for(auto &x: nums) {
            x += base;
        }
        int cur = 1;
        for(int i = 0; i <= 6; ++ i) {
            for(int j = 0; j < 10; ++ j) {
                cnt[j].clear();
            }
            for(auto x: nums) {
                cnt[(x/cur) % 10].push_back(x);
            }
            nums.clear();
            for(int j = 0; j < 10; ++ j) {
                for(auto x: cnt[j]) {
                    nums.push_back(x);
                }
            }
            cur *= 10;
        }
        for(auto &x: nums) x -= base;
        return nums;
    }
};
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