How Many Numbers Are Smaller Than the Current Number

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

 

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation: 
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). 
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1). 
For nums[3]=2 there exist one smaller number than it (1). 
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

 

Constraints:

  • 2 <= nums.length <= 500
  • 0 <= nums[i] <= 100
SOLUTION:
import bisect

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        n = len(nums)
        ans = [0] * n
        order = sorted([(num, i) for i, num in enumerate(nums)])
        for i in range(n):
            pos = bisect.bisect_left(order, (order[i][0], float('-inf')))
            ans[order[i][1]] = i - max(i - pos, 0)
        return ans

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