LeetCode in Kotlin
1365. How Many Numbers Are Smaller Than the Current Number
Easy
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 <= 5000 <= nums[i] <= 100
Solution
class Solution {
fun smallerNumbersThanCurrent(nums: IntArray): IntArray {
val ans = IntArray(nums.size)
val temp = IntArray(101)
for (num in nums) {
temp[num]++
}
for (i in 1..100) {
temp[i] += temp[i - 1]
}
for (i in ans.indices) {
if (nums[i] == 0) {
ans[i] = 0
} else {
ans[i] = temp[nums[i] - 1]
}
}
return ans
}
}