LeetCode in Kotlin
1574. Shortest Subarray to be Removed to Make Array Sorted
Medium
Given an integer array arr, remove a subarray (can be empty) from arr such that the remaining elements in arr are non-decreasing.
Return the length of the shortest subarray to remove.
A subarray is a contiguous subsequence of the array.
Example 1:
Input: arr = [1,2,3,10,4,2,3,5]
Output: 3
Explanation: The shortest subarray we can remove is [10,4,2] of length 3. The remaining elements after that will be [1,2,3,3,5] which are sorted. Another correct solution is to remove the subarray [3,10,4].
Example 2:
Input: arr = [5,4,3,2,1]
Output: 4
Explanation: Since the array is strictly decreasing, we can only keep a single element. Therefore we need to remove a subarray of length 4, either [5,4,3,2] or [4,3,2,1].
Example 3:
Input: arr = [1,2,3]
Output: 0
Explanation: The array is already non-decreasing. We do not need to remove any elements.
Constraints:
1 <= arr.length <= 1050 <= arr[i] <= 109
Solution
class Solution {
fun findLengthOfShortestSubarray(arr: IntArray): Int {
var left = 0
while (left < arr.size - 1 && arr[left] <= arr[left + 1]) {
left++
}
if (left == arr.size - 1) {
return 0
}
var right = arr.size - 1
while (right > left && arr[right] >= arr[right - 1]) {
right--
}
if (right == 0) {
return arr.size - 1
}
var result = Math.min(arr.size - left - 1, right)
var i = 0
var j = right
while (i <= left && j < arr.size) {
if (arr[j] >= arr[i]) {
result = Math.min(result, j - i - 1)
i++
} else {
j++
}
}
return result
}
}