<p>You are given an integer array <code>nums</code>. You need to create a 2D array from <code>nums</code> satisfying the following conditions:</p>

<ul>
	<li>The 2D array should contain <strong>only</strong> the elements of the array <code>nums</code>.</li>
	<li>Each row in the 2D array contains <strong>distinct</strong> integers.</li>
	<li>The number of rows in the 2D array should be <strong>minimal</strong>.</li>
</ul>

<p>Return <em>the resulting array</em>. If there are multiple answers, return any of them.</p>

<p><strong>Note</strong> that the 2D array can have a different number of elements on each row.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,4,1,2,3,1]
<strong>Output:</strong> [[1,3,4,2],[1,3],[1]]
<strong>Explanation:</strong> We can create a 2D array that contains the following rows:
- 1,3,4,2
- 1,3
- 1
All elements of nums were used, and each row of the 2D array contains distinct integers, so it is a valid answer.
It can be shown that we cannot have less than 3 rows in a valid array.</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> [[4,3,2,1]]
<strong>Explanation:</strong> All elements of the array are distinct, so we can keep all of them in the first row of the 2D array.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>1 &lt;= nums[i] &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> findMatrix(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int i, j = 0, f = 1, mf = 1;
        for (i = 1 ; i < nums.size() ; i++)
            if (nums[i - 1] == nums[i]) f++;
            else{
                if (mf < f) mf = f;
                f = 1;        
            }
        if (mf < f) mf = f;
        vector<vector<int>> r(mf);
        r[j].push_back(nums[0]);
        for (i = 1 ; i < nums.size() ; i++)
            if (nums[i - 1] == nums[i]) r[++j].push_back(nums[i]);
            else{
                if (j > 0)  j = 0;
                r[j].push_back(nums[i]);
            }
        return r;
    }
};
32800