[LeetCode]#1365. How Many Numbers Are Smaller Than the Current Number

Environment: Python 3.7

Key technique: for loop, if

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

Analysis:

1. Use target number as 8 and subtract all other number.
2. Count all less than zero number.

Solution:

class Solution:
    def smallerNumbersThanCurrent(self, nums):
        l=len(nums)
        ans=[]
        count=0
        for i in nums:
            for j in range(l):
               if (nums[j]-i)<0:
                   count+=1
            ans.append(count)
            count=0
        return ans

Submitted result: