Given an integer array arr, remove a subarray (can be empty) from arr such that the remaining elements in arr are non-decreasing.
A subarray is a contiguous subsequence of the array.
Return the length of the shortest subarray to remove.
 
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.

Example 4:
Input: arr = [1]
Output: 0

 
Constraints:

1 <= arr.length <= 10^5
0 <= arr[i] <= 10^9

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

def _disabled_input(*args, **kwargs):
    raise RuntimeError("input() is disabled in callable mode")
builtins.input = _disabled_input

ns = {"__name__": "__judge__"}  # prevent if __name__ == "__main__" blocks
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'class Solution:\n    def findLengthOfShortestSubarray(self, arr: List[int]) -> int:\n        #https://leetcode.com/problems/shortest-subarray-to-be-removed-to-make-array-sorted/discuss/830480/C%2B%2B-O(N)-Sliding-window-Explanation-with-Illustrations\n        #这道题是two pointer类型题目\n        #大体的逻辑就是左边一个pointer，右边一个pointer\n        left = 0\n        right = len(arr) - 1\n        \n        #left向右走，如果left+1指向的数字大于等于left就一直走\n        #如果left一路走到了end，那就说明原list就是sorted的，那就直接return\n        while left + 1 <= len(arr) - 1 and arr[left] <= arr[left+1]:\n            left += 1\n        \n        if left == right:\n            return 0\n        \n        #right同理向左走，如果right-1小于等于right就继续走\n        while right - 1 >= 0 and arr[right] >= arr[right - 1]:\n            right -= 1\n            \n        #然后先处理两边 就是当0到right-1 或者 left+1到end 其中一个去掉，也就是去掉两边\n        #在这种情况下 无论怎么移除中间的部分 都没有办法使两边组合起来成为递增\n        #这种情况就像[6,7,8,9,10,3,5,20,4,1,2,3]\n        #left指向10，right指向1 -- 因为只能移除一个subarry所以要从 6到20 和 3到3 中选一个\n        \n        \n        #所以如果是这种case，那就先要计算两个边较小的那个\n        #right就是0到right -1 index的长度\n        #len(arr)- 1 - left 就是 end - left 也就是 left + 1到end index的长度  \n        min_sublen = min(right, len(arr)- 1 - left)\n        \n        #然后就可以开始进行中间情况的运算\n        #大概就是在这种情况下我们left和right要进行移动，然后找到最短的中间subarray，使得该array去掉后左右两边能成为递增array\n        #大概的操作就是比较 两个指针lp和rp 从0到left 和 从right到end\n       \n        #如果lp小于rp了，那就说明去掉lp和rp中间范围就是递增，此时就可以计算下subarr长度，然后进行比较，留下最小的那个。然后lp向右移动一位再继续比较\n        #知道lp到达了left，或者rp到达了end，就停止 返回最短的subarry长度\n        lp = 0\n        rp = right\n        \n        #如果lp大于rp，那就说明lp太大了，rp向右移动一位\n        while lp <= left and rp <= len(arr) - 1:\n            if arr[lp] > arr[rp]:\n                rp += 1\n            #如果lp小于rp了，那就说明去掉lp和rp中间范围就是递增，此时就可以计算下subarr长度，然后进行比较，留下最小的那个。然后lp向右移动一位再继续比较\n            else:\n                min_sublen = min(min_sublen, rp-lp-1)\n                lp += 1\n        \n        return min_sublen'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'findLengthOfShortestSubarray'
if _fn_name:
    if "Solution" in ns:
        try:
            _obj = ns["Solution"]()
            if hasattr(_obj, _fn_name):
                _target = getattr(_obj, _fn_name)
        except Exception:
            _target = None
    if _target is None and _fn_name in ns and callable(ns[_fn_name]):
        _target = ns[_fn_name]
# Fallback: if there's exactly one user-defined function, use it
if _target is None:
    _cands = [v for k, v in ns.items() if callable(v) and getattr(v, "__module__", "") in ("__judge__", "__main__")]
    _user_funcs = [c for c in _cands if getattr(c, "__name__", "").startswith("_") is False]
    if len(_user_funcs) == 1:
        _target = _user_funcs[0]
if _target is None:
    raise RuntimeError(f"Could not resolve callable '{_fn_name}'")

_raw = sys.stdin.read().strip()
if _raw:
    _parsed = json.loads(_raw)
    if isinstance(_parsed, dict):
        _kwargs = _parsed.get("kwargs", {})
        _args = _parsed.get("args", [])
    else:
        _args, _kwargs = _parsed, {}
else:
    _args, _kwargs = [], {}
if not isinstance(_args, (list, tuple)):
    _args = [_args]
_res = _target(*_args, **_kwargs)
try:
    sys.stdout.write(repr(_res) + "\n")
except Exception:
    pass

```

Measured memory_bytes: 5938