# 164 Maximum Gap

## 164. [Maximum Gap](https://leetcode.com/problems/maximum-gap/description/)

## 1. Question

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

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

## 2. Implementation

**(1) Sort**

```java
class Solution {
    public int maximumGap(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }

        Arrays.sort(nums);
        int res = 0;

        for (int i = 0; i < nums.length - 1; i++) {
            res = Math.max(res, nums[i + 1] - nums[i]);
        }
        return res;
    }
}
```

**(2) Bucket Sort**

```java
class Solution {
    public int maximumGap(int[] nums) {
        if (nums == null || nums.length < 2) {
            return 0;
        }

        int n = nums.length;
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;

        for (int num : nums) {
            min = Math.min(min, num);
            max = Math.max(max, num);
        }

        if (min == max) {
            return 0;
        }

        Bucket buckets[] = new Bucket[n];

        for (int i = 0; i < n; i++) {
            buckets[i] = new Bucket();
        }

        int bucketLen = (int)Math.ceil((double)(max - min) / (n - 1));

        for (int num : nums) {
            int index = (num - min)/bucketLen;

            if (buckets[index].min == Integer.MAX_VALUE) {
                buckets[index].min = num;
                buckets[index].max = num;
            }
            else {
                buckets[index].min = Math.min(buckets[index].min, num);
                buckets[index].max = Math.max(buckets[index].max, num);
            }
        }

        int maxGap = 0, prev = 0;
        for (int i = 1; i < n; i++) {
            if (buckets[i].min != Integer.MAX_VALUE) {
                maxGap = Math.max(maxGap, buckets[i].min - buckets[prev].max);
                prev = i;
            }
        }
        return maxGap;
    }

    class Bucket {
        int min, max;

        public Bucket() {
            min = Integer.MAX_VALUE;
            max = Integer.MIN_VALUE;
        }
    }
}
```

## 3. Time & Space Complexity

Sort: 时间复杂度O(nlogn), 空间复杂度O(1)

Bucket Sort: 时间复杂度O(n), 空间复杂度O(n)


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