Leetcode - How Many Numbers Are Smaller Than the Current Number Solution
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.
Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums=1 does not exist any smaller number than it. For nums=2 there exist one smaller number than it (1). For nums=2 there exist one smaller number than it (1). For nums=3 there exist three smaller numbers than it (1, 2 and 2).
Input: nums = [6,5,4,8] Output: [2,1,0,3]
Input: nums = [7,7,7,7] Output: [0,0,0,0]
2 <= nums.length <= 500
0 <= nums[i] <= 100
Solution in Python
from collections import Counter class Solution: def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]: counts = Counter(nums) smaller_nums =  total = 0 for x,y in sorted(counts.items()): total += y smaller_nums.append((x,total-y)) smaller_nums = dict(smaller_nums) return [smaller_nums[i] for i in nums]