# 1574. Shortest Subarray to be Removed to Make Array Sorted

## Description

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 <= 105
• 0 <= arr[i] <= 109

## Solutions

• class Solution {
public int findLengthOfShortestSubarray(int[] arr) {
int n = arr.length;
int i = 0, j = n - 1;
while (i + 1 < n && arr[i] <= arr[i + 1]) {
++i;
}
while (j - 1 >= 0 && arr[j - 1] <= arr[j]) {
--j;
}
if (i >= j) {
return 0;
}
int ans = Math.min(n - i - 1, j);
for (int l = 0; l <= i; ++l) {
int r = search(arr, arr[l], j);
ans = Math.min(ans, r - l - 1);
}
return ans;
}

private int search(int[] arr, int x, int left) {
int right = arr.length;
while (left < right) {
int mid = (left + right) >> 1;
if (arr[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}

• class Solution {
public:
int findLengthOfShortestSubarray(vector<int>& arr) {
int n = arr.size();
int i = 0, j = n - 1;
while (i + 1 < n && arr[i] <= arr[i + 1]) {
++i;
}
while (j - 1 >= 0 && arr[j - 1] <= arr[j]) {
--j;
}
if (i >= j) {
return 0;
}
int ans = min(n - 1 - i, j);
for (int l = 0; l <= i; ++l) {
int r = lower_bound(arr.begin() + j, arr.end(), arr[l]) - arr.begin();
ans = min(ans, r - l - 1);
}
return ans;
}
};

• class Solution:
def findLengthOfShortestSubarray(self, arr: List[int]) -> int:
n = len(arr)
i, j = 0, n - 1
while i + 1 < n and arr[i] <= arr[i + 1]:
i += 1
while j - 1 >= 0 and arr[j - 1] <= arr[j]:
j -= 1
if i >= j:
return 0
ans = min(n - i - 1, j)
for l in range(i + 1):
r = bisect_left(arr, arr[l], lo=j)
ans = min(ans, r - l - 1)
return ans


• func findLengthOfShortestSubarray(arr []int) int {
n := len(arr)
i, j := 0, n-1
for i+1 < n && arr[i] <= arr[i+1] {
i++
}
for j-1 >= 0 && arr[j-1] <= arr[j] {
j--
}
if i >= j {
return 0
}
ans := min(n-i-1, j)
for l := 0; l <= i; l++ {
r := j + sort.SearchInts(arr[j:], arr[l])
ans = min(ans, r-l-1)
}
return ans
}