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Abhishek Chaudhary

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# 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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