<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>

#include <ranges>

class Solution
{
    public:
    int maximumGap(std::vector<int>& nums) const
    {
        if(nums.size()<2) return 0;
        
        std::array<std::vector<int>,10> buckets;

        for(unsigned mod = 1; mod<=1000000000u; mod*=10)
        {
            for(auto v: nums)
                buckets[(v/mod)%10].push_back(v);

            for(std::size_t idx = 0; const auto& bucket: buckets)
                for(auto v: bucket)
                    nums[idx++] = v;
                
            for(auto& bucket: buckets)
                bucket.clear();
        }

        std::adjacent_difference(nums.begin(),nums.end(),nums.begin(), [](int a, int b)
        {
            return std::abs(a-b);
        });
        return *std::max_element(std::next(nums.begin()),nums.end());
    }
};
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