<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) {
        vector<int> idx(nums.size());
        for (int i = 0; i < nums.size(); ++i) idx[i] = i;

        int min = *min_element(nums.begin(), nums.end());

        for (int iter = 0; iter < 32; ++iter) {
            vector<vector<int>> buckets(2);

            for (int i : idx) {
                int v = nums[i] - min;
                int b = (v >> iter) & 1;
                buckets[b].push_back(i);
            }

            int next_i = 0;
            for (auto &bu : buckets) {
                for (auto i : bu) {
                    idx[next_i] = i;
                    ++next_i;
                }
            }
        }

        vector<int> ans;
        for (auto i : idx) ans.push_back(nums[i]);
        return ans;

    }

   
};
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