1365-how-many-numbers-are-smaller-than-the-current-number¶

Description¶

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(Python)¶

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        return  self.sorting(nums)
    def sorting(self, nums: List[int]) -> List[int]:
        res  = []
        # sort the array using bucket sort
        bucket = [0] * (max(nums) + 1)
        for num in nums:
            bucket[num] += 1 # [0 1 2 1 0 0 0 0 1]
        for i in range(1, len(bucket)):
            bucket[i] += bucket[i-1] # [0 1 3 4 4 4 4 4 5]
        for num in nums:
            if num == 0:
                res.append(0)
            else:
                res.append(bucket[num-1])
        return res
            
        