<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) {
        if (nums.size() < 2) return 0;
        radixSort(nums);
        
        int max_gap = 0;
        for (size_t i = 1; i < nums.size(); ++i) {
            max_gap = max(max_gap, nums[i] - nums[i - 1]);
        }
        
        return max_gap;
    }
    
private:
    void radixSort(vector<int>& nums) {
        int max_num = *max_element(nums.begin(), nums.end());
        int exp = 1;  // Initial exponent (1s place)
        
        while (max_num / exp > 0) {
            countingSortByDigit(nums, exp);
            exp *= 10;
        }
    }
    
    void countingSortByDigit(vector<int>& nums, int exp) {
        vector<int> output(nums.size());
        vector<int> count(10, 0);
        for (int num : nums) {
            count[(num / exp) % 10]++;
        }
        
        for (int i = 1; i < 10; ++i) {
            count[i] += count[i - 1];
        }
        
        for (int i = nums.size() - 1; i >= 0; --i) {
            output[count[(nums[i] / exp) % 10] - 1] = nums[i];
            count[(nums[i] / exp) % 10]--;
        }
        
        for (size_t i = 0; i < nums.size(); ++i) {
            nums[i] = output[i];
        }
    }
};

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