Shortest Subarray to be Removed to Make Array Sorted
<p>Given an integer array <code>arr</code>, remove a subarray (can be empty) from <code>arr</code> such that the remaining elements in <code>arr</code> are <strong>non-decreasing</strong>.</p>

<p>Return <em>the length of the shortest subarray to remove</em>.</p>

<p>A <strong>subarray</strong> is a contiguous subsequence of the array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3,10,4,2,3,5]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [5,4,3,2,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,2,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The array is already non-decreasing. We do not need to remove any elements.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Stack; Array; Two Pointers; Binary Search; Monotonic Stack
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;
    }
};