<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:
    def maximumGap(self, nums: List[int]) -> int:
        n=len(str(max(nums)))
        nums=[str(i) for i in nums]
        for i in range(len(nums)):
            nums[i] = '0' * (n - len(nums[i])) + nums[i] 
        for i in range(n-1,-1,-1):
            li=[0]*10
            for j in nums:
                li[int(j[i])]+=1
            for k in range(1,10):
                li[k]+=li[k-1]
            res=[0]*len(nums)
            for k in reversed(nums):
                res[li[int(k[i])]-1]=k
                li[int(k[i])]-=1
            nums=res
        nums=[int(i) for i in nums]
        m=0
        for i in range(1,len(nums)):
            m=max(m,nums[i]-nums[i-1])
        return (m)
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