LeetCode in Kotlin
2089. Find Target Indices After Sorting Array
Easy
You are given a 0-indexed integer array nums and a target element target.
A target index is an index i such that nums[i] == target.
Return a list of the target indices of nums after sorting nums in non-decreasing order. If there are no target indices, return an empty list. The returned list must be sorted in increasing order.
Example 1:
Input: nums = [1,2,5,2,3], target = 2
Output: [1,2]
Explanation: After sorting, nums is [1,2,2,3,5].
The indices where nums[i] == 2 are 1 and 2.
Example 2:
Input: nums = [1,2,5,2,3], target = 3
Output: [3]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 3 is 3.
Example 3:
Input: nums = [1,2,5,2,3], target = 5
Output: [4]
Explanation: After sorting, nums is [1,2,2,3,5].
The index where nums[i] == 5 is 4.
Constraints:
1 <= nums.length <= 1001 <= nums[i], target <= 100
Solution
class Solution {
fun targetIndices(nums: IntArray, target: Int): List<Int> {
var count = 0
var lessthan = 0
for (n in nums) {
if (n == target) {
count++
}
if (n < target) {
lessthan++
}
}
val result: MutableList<Int> = ArrayList()
for (i in 0 until count) {
result.add(lessthan++)
}
return result
}
}