1365. How Many Numbers Are Smaller Than the Current Number.js

// 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]

const smallerNumbersThanCurrent = nums => {
	const sorted = [...nums].sort((a, b) => a - b)
	return nums.map(num => sorted.indexOf(num))
}