# 164. Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

Input: [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either
             (3,6) or (6,9) has the maximum difference 3.

Example 2:

Input: [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.

Note:

  • You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.
  • Try to solve it in linear time/space.

### Solution

Approach 1: Sorting.

Approach 2: Bucket sort.

Approach 3: Radix sort.

### Code (Python)

Approach 1:
```python

Approach 2:

# Code (C++)

Approach 1:

// Sorting.
class Solution {
public:
    int maximumGap(vector<int>& nums) {
        std::sort(nums.begin(), nums.end());
        int maxGap = 0;
        for (int i = 1; i < nums.size(); ++i)
        {
            maxGap = std::max(maxGap, nums[i] - nums[i-1]);
        }
        return maxGap;
    }
};

Approach 2:

// Bucket sort.
class Solution {
public:
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2)
            return 0;
        int max = nums[0];
        int min = nums[0];
        for (int i = 1; i < n; ++i)
        {
            max = std::max(max, nums[i]);
            min = std::min(min, nums[i]);
        }
        vector<int> bucketMax(n, min-1);
        vector<int> bucketMin(n, max+1);
        int interval = (max - min + 1) / n;
        if (interval == 0 || (max - min + 1) % n != 0)
            interval++;
        for (int i = 0; i < n; ++i)
        {
            int id = (nums[i] - min) / interval;
            bucketMax[id] = std::max(bucketMax[id], nums[i]);
            bucketMin[id] = std::min(bucketMin[id], nums[i]);
        }
        int maxGap = 0;
        int prevMax = bucketMax[0];
        for (int i = 1; i < n; ++i)
        {
            if (bucketMin[i] <= max)
            {
                maxGap = std::max(maxGap, bucketMin[i] - prevMax);
                prevMax = bucketMax[i];
            }
        }
        return maxGap;
    }
};

Approach 3:

// radix sort
class Solution {
public:
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2)
            return 0;
        for (int bit = 0; bit < 32; ++bit) {
            std::stable_partition(nums.begin(), nums.end(),
                [bit](int a) {
                    return !(a & (1 << bit));
                });
        }
        int maxGap = -1;
        for (int i = 1; i < n; ++i) {
            maxGap = std::max(maxGap, nums[i] - nums[i-1]);
        }
        return maxGap;
    }
};