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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/how-many-numbers-are-smaller-than-the-current-number
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# version 1
class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        res = []
        for i in range(len(nums)):
            count = 0
            for j in range(len(nums)):                
                if nums[j] < nums[i]:
                    count += 1
            res.append(count)
        return res

# version 2
class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        # 复制 nums 到 nums1 有两个方法
        # nums1 = [x for x in nums]  # 方法1
        nums1 = copy.deepcopy(nums) # 方法2
        nums1.sort() # 排序之后，数字的下标就是有多少数字小于它
        res = []
        for num in nums:
            res.append(nums1.index(num))
        return res