<p>Given an integer array <code>nums</code>, return <em>the maximum difference between two successive elements in its sorted form</em>. If the array contains less than two elements, return <code>0</code>.</p>

<p>You must write an algorithm that runs in linear time and uses linear extra space.</p>

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

<pre>
<strong>Input:</strong> nums = [3,6,9,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The array contains less than 2 elements, therefore return 0.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int getChar(int num, long k) // 1941 - k = 100 -> get 4
    {
        num -= num % (k / 10);
        num = num % (k);
        num = num / (k / 10);
        return num;
    }
    int maximumGap(vector<int>& nums) {
        int d = 10;
        int n = nums.size();
        long k = 1;
        vector<queue<int>> arr(10);
        int index = 0;
        for(int i = 0; i < d; ++i)
        {
            k *= 10;
            for(int j = 0; j < n; ++j)
            {
                arr[getChar(nums[j], k)].push(nums[j]);
            }
            index = 0;
            for(int j = 0; j < 10; ++j)
            {
                while(arr[j].size())
                {
                    nums[index++] = arr[j].front();
                    arr[j].pop();
                }
            }
        }
        int res = 0;
        for(int i = 1; i < n; ++i)
            res = max(res, nums[i] - nums[i-1]);
        return res;
    }
};
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