LeetCode 315. Count of Smaller Numbers After Self

Description

https://leetcode.com/problems/count-of-smaller-numbers-after-self/

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 1:

Input: nums = [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.

Constraints:

  • 0 <= nums.length <= 10^5
  • -10^4 <= nums[i] <= 10^4

Explanation

Python Solution

class Solution:
    def countSmaller(self, nums):
        result = []
        
        sorted_nums = []
        
        for num in reversed(nums):
            index = bisect.bisect_left(sorted_nums, num)
            result.insert(0, index)
            sorted_nums.insert(index, num)
                    
        return result
    
  • Time Complexity: ~Nlog(N)
  • Space Complexity: ~N.

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