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

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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]:
        # nums 8 1 2 2 3
        # 2 -> 1
        # x > nums[j]
        # i -> left .... i...right
        # greatest_from_left = 0 0 0 0 0
        #                      8 1 2 2 3 
        #                      8 > 1
        #                      0 1 0 0 0
        #                      2 
        # greatest_from_right = 0 0 0 0 0
        # [4,0,1,1,3]
        # 
        # 8 1 2 2 3
        # [0 1 2 1 0 0 0 0 1]
        # [0 1 3 4 4 4 4 4 5]
        # since strictly smaller look cum[num-1]
        n = max(nums) + 1
        count_sum = [0] * n
        for num in nums:
            count_sum[num] += 1
        
        for i in range(1, n):
            count_sum[i] += count_sum[i-1]

        return [0 if num == 0  else count_sum[num-1]  for num in nums]

Dry Run