<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) {

        for (int i = 0; i < 4; i++) {

            sortByByte(i, nums);

        }

        int ans = 0;

        for (int i = 1; i < nums.size(); i++) {
            ans = max(ans, nums[i] - nums[i - 1]);
        }

        return ans;
        
    }

    void sortByByte(int theByte, vector<int>& nums) {

        vector<int> buckets[256];

        for (auto num : nums) {

            int bucket = getBucket(theByte, num);
            buckets[bucket].push_back(num);

        }

        nums.clear();

        for (int i = 0; i < 256; i++) {

            for (auto num : buckets[i]) {
                nums.push_back(num);
            }

        }

    }

    int getBucket(int theByte, int num) {
        
        int mask = 0x000000FF;

        for (int i = 0; i < theByte; i++) {
            num >>= 8;
        }

        return mask & num;
    }
};
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