LeetCode： How Many Numbers Are Smaller Than the Current Number

Question

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]

Solution #1

def smaller_numbers_than_current(nums)
  dict = Hash.new(0)
  j = 0
  result = nums
  nums = nums.sort.reverse

  nums.each do |num|
    dict[num] = 0
    while j < nums.length - 1
      dict[num] += 1 if nums[j + 1] && num > nums[j + 1]
      j += 1
    end
    j = 0
  end
  result.map { |num| dict[num] }
end

Solution #2

def smaller_numbers_than_current(nums)
  result = nums.dup
  dict = Hash.new(0)

  nums.sort!.map { |num| dict[num] = nums.index(num) }
  result.map { |num| dict[num] }
end