LeetCode 164 - Maximum Gap (Hard)

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.

 方法1： 按照题意直接做即可。

思路：先排序。然后比较得出最大的max gap

time complexity： O(nlogn) space complexity O(1） 

class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        if not nums or len(nums) < 2: return 0 
        
        nums.sort() 
        res = 0 
        for i in range(1, len(nums)):
            res = max(res, nums[i] - nums[i-1])
        return res 

方法2： bucket sort

思路：

1 找到整个list的最大值max 和最小值min

2 得到bucket size => （max-min)//(n - 1） （不减1也可以）

3 得到buckNum => (max - min ) // bucket size + 1 （at least需要一个桶，所以一定要+1）

4 把每个list中的数字放到相应的bucket中。筛掉没有使用过的桶。最后在有效的桶中，遍历一遍即可。

time complexity：O(n)  space  complexity: O(b) b is the number of buckets 

class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        if not nums or len(nums) < 2: return 0 
        
        minNum = min(nums)
        maxNum = max(nums)
        
        n = len(nums)
        buckSize = max(1, (maxNum - minNum) // (n - 1))
        buckNum = (maxNum - minNum) // buckSize + 1 
        
        buckets = [[None, None] for _ in range(buckNum)]
     
        for num in nums:
            index = (num - minNum) // buckSize 
            buckets[index][0] = num if buckets[index][0] is None else min(num, buckets[index][0])
            buckets[index][1] = num if buckets[index][1] is None else max(num, buckets[index][1])
            
        validBucket = [b for b in buckets if b[0] is not None]
        
        res = 0 
        for i in range(1, len(validBucket)):
            res = max(res, validBucket[i][0] - validBucket[i-1][1])
            
        return res