<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 maximumGap(vector<int>& nums) {
        int size = nums.size();
        if(size < 2)
           return 0;
        int max_el = nums[0];
        for(int i = 1; i < size; ++i)
        {
            if(nums[i] >  max_el)
               max_el = nums[i];
        }
        
        vector<int> output(size,0);
        for(int exp = 1; max_el/exp > 0; exp *= 10)
        {
            vector<int> index(10,0);
            for(int i = 0; i < size; ++i)
            {
                ++index[(nums[i]/exp%10)];
            }
            for(int i = 1; i < 10; ++i)
            {
                index[i] += index[i-1];
            }
            for(int i = size - 1 ; i >= 0; --i)
            {
                int temp = nums[i]/exp%10;
                output[index[temp]-1] =  nums[i];
                --index[temp];
            }
            for(int i = 0; i < size; ++i)
            {
                nums[i] = output[i];
            }


        }
        int max = 0;
        for(int i = 1; i < size; ++i)
        {
            if(output[i] - output[i-1] > max)
               max = output[i] - output[i-1];
        }
        return max;
        
    }
};
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