LeetCode 315. Count of Smaller Numbers After Self 归并排序

最新推荐文章于 2024-07-07 11:07:39 发布

原创最新推荐文章于 2024-07-07 11:07:39 发布 · 216 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

475 篇文章

订阅专栏

leetcode

231 篇文章

订阅专栏

本文介绍了一种解决计数右侧更小元素问题的方法，即返回一个数组，其中每个元素表示原数组中该位置右侧有多少个比它小的元素。通过使用深度优先搜索(DFS)的分治策略，实现了一个时间复杂度为O(nlogn)的算法。

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example:

Input: [5,2,6,1]
Output: [2,1,1,0] 
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

------------------------------------------------------------------------------

The right code is :

class Solution:
    def dfs(self, nums, res, start, end):
        if (start > end):
            return []
        elif (start == end):
            return [nums[start]]

        mid = start + ((end - start) >> 1)
        left = self.dfs(nums, res, start, mid)
        right = self.dfs(nums, res, mid + 1, end)
        sorted_tuples = [(0, 0) for i in range(end - start + 1)]
        for i in range(end - start + 1)[::-1]:
            if ((not right) or (left and left[-1][1] > right[-1][1])):
                res[left[-1][0]] += len(right)
                sorted_tuples[i] = left.pop()
            else:
                sorted_tuples[i] = right.pop()
        return sorted_tuples

    def countSmaller(self, nums):
        l = len(nums)
        res = [0 for i in range(l)]
        self.dfs(list(enumerate(nums)), res, 0, l - 1)
        return res

For merge sort, the time complexity is O(nlogn). O(logn) is the tree height, O(n) is merging complexity. I wrote an O(n^2logn) complexity firstly.