LeetCode in Kotlin
1673. Find the Most Competitive Subsequence
Medium
Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k.
An array’s subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 is less than 5.
Example 1:
Input: nums = [3,5,2,6], k = 2
Output: [2,6]
Explanation: Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.
Example 2:
Input: nums = [2,4,3,3,5,4,9,6], k = 4
Output: [2,3,3,4]
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1091 <= k <= nums.length
Solution
class Solution {
fun mostCompetitive(nums: IntArray, k: Int): IntArray {
val r = IntArray(k)
val n = nums.size
var j = 0
for (i in 0 until n) {
if (i == 0) {
r[j] = nums[i]
j++
} else {
var l = j - 1
while (l >= 0 && nums[i] < r[l] && n - i >= k - l) {
l--
}
j = l + 1
if (j < k) {
r[j] = nums[i]
j++
}
}
}
return r
}
}