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Abhishek Chaudhary
Abhishek Chaudhary

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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.

Example 2:

Input: nums = [-1]
Output: [0]

Example 3:

Input: nums = [-1,-1]
Output: [0,0]

Constraints:

  • 1 <= nums.length <= 105
  • -104 <= nums[i] <= 104

SOLUTION:

import bisect

class Solution:
    def countSmaller(self, nums: List[int]) -> List[int]:
        n = len(nums)
        smallers = []
        for i in range(n - 1, -1, -1):
            currctr = bisect.bisect_left(smallers, nums[i])
            bisect.insort(smallers, nums[i])
            nums[i] = currctr
        return nums
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