letcode_1365.py

# 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).

nums = [8,1,2,2,3]

resultado = []
cont= 0
for i in nums:
    for e in range(len(nums)):
        
        if i > nums[e]:
            cont += 1
    resultado.append(cont)
    cont = 0
print(resultado)