1365. How Many Numbers Are Smaller Than the Current Number (Easy) (https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number)
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. Constraints: - 2 <= nums.length <= 500 - 0 <= nums[i] <= 100
function smallerNumbersThanCurrent(nums: number[]): number[] {
const sorted = [...nums].sort((a, b) => a - b)
const firstIndex = new Map<number, number>()
for (let i = 0; i < sorted.length; i++) {
if (!firstIndex.has(sorted[i])) {
firstIndex.set(sorted[i], i)
}
}
return nums.map(n => firstIndex.get(n)!)
}
// Local check:
console.log(smallerNumbersThanCurrent([8, 1, 2, 2, 3]))
console.log(smallerNumbersThanCurrent([6, 5, 4, 8]))
console.log(smallerNumbersThanCurrent([7, 7, 7, 7]))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]