Make It Good
Contest: Codeforces Round 656 (Div. 3)
Difficulty rating: 1200
Tags: greedy

You are given an array $$$a$$$ consisting of $$$n$$$ integers. You have to find the length of the smallest (shortest) prefix of elements you need to erase from $$$a$$$ to make it a good array. Recall that the prefix of the array $$$a=[a_1, a_2, \dots, a_n]$$$ is a subarray consisting several first elements: the prefix of the array $$$a$$$ of length $$$k$$$ is the array $$$[a_1, a_2, \dots, a_k]$$$ ($$$0 \le k \le n$$$).

The array $$$b$$$ of length $$$m$$$ is called good, if you can obtain a non-decreasing array $$$c$$$ ($$$c_1 \le c_2 \le \dots \le c_{m}$$$) from it, repeating the following operation $$$m$$$ times (initially, $$$c$$$ is empty):

- select either the first or the last element of $$$b$$$, remove it from $$$b$$$, and append it to the end of the array $$$c$$$.

For example, if we do $$$4$$$ operations: take $$$b_1$$$, then $$$b_{m}$$$, then $$$b_{m-1}$$$ and at last $$$b_2$$$, then $$$b$$$ becomes $$$[b_3, b_4, \dots, b_{m-3}]$$$ and $$$c =[b_1, b_{m}, b_{m-1}, b_2]$$$.

Consider the following example: $$$b = [1, 2, 3, 4, 4, 2, 1]$$$. This array is good because we can obtain non-decreasing array $$$c$$$ from it by the following sequence of operations:

1. take the first element of $$$b$$$, so $$$b = [2, 3, 4, 4, 2, 1]$$$, $$$c = [1]$$$;
2. take the last element of $$$b$$$, so $$$b = [2, 3, 4, 4, 2]$$$, $$$c = [1, 1]$$$;
3. take the last element of $$$b$$$, so $$$b = [2, 3, 4, 4]$$$, $$$c = [1, 1, 2]$$$;
4. take the first element of $$$b$$$, so $$$b = [3, 4, 4]$$$, $$$c = [1, 1, 2, 2]$$$;
5. take the first element of $$$b$$$, so $$$b = [4, 4]$$$, $$$c = [1, 1, 2, 2, 3]$$$;
6. take the last element of $$$b$$$, so $$$b = [4]$$$, $$$c = [1, 1, 2, 2, 3, 4]$$$;
7. take the only element of $$$b$$$, so $$$b = []$$$, $$$c = [1, 1, 2, 2, 3, 4, 4]$$$ — $$$c$$$ is non-decreasing.

Note that the array consisting of one element is good.

Print the length of the shortest prefix of $$$a$$$ to delete (erase), to make $$$a$$$ to be a good array. Note that the required length can be $$$0$$$.

You have to answer $$$t$$$ independent test cases.

Input format:
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 2 \cdot 10^4$$$) — the number of test cases. Then $$$t$$$ test cases follow.

The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of $$$a$$$. The second line of the test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 2 \cdot 10^5$$$), where $$$a_i$$$ is the $$$i$$$-th element of $$$a$$$.

It is guaranteed that the sum of $$$n$$$ does not exceed $$$2 \cdot 10^5$$$ ($$$\sum n \le 2 \cdot 10^5$$$).

Output format:
For each test case, print the answer: the length of the shortest prefix of elements you need to erase from $$$a$$$ to make it a good array.

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case of the example, the array $$$a$$$ is already good, so we don't need to erase any prefix.

In the second test case of the example, the initial array $$$a$$$ is not good. Let's erase first $$$4$$$ elements of $$$a$$$, the result is $$$[4, 5, 2]$$$. The resulting array is good. You can prove that if you erase fewer number of first elements, the result will not be good.

Example input:
5
4
1 2 3 4
7
4 3 3 8 4 5 2
3
1 1 1
7
1 3 1 4 5 3 2
5
5 4 3 2 3

Example output:
0
4
0
2
3

Editorial:
Consider the maximum element $$$a_{mx}$$$ of the good array $$$a$$$ of length $$$k$$$. Then we can notice that the array $$$a$$$ looks like $$$[a_1 \le a_2 \le \dots \le a_{mx} \ge \dots \ge a_{k-1} \ge a_k]$$$. And this is pretty obvious that if the array doesn't have this structure, then it isn't good (you can see it yourself). So we need to find the longest such suffix. It's pretty easy doable with pointer: initially, the pointer $$$pos$$$ is at the last element. Then, while $$$pos > 1$$$ and $$$a_{pos - 1} \ge a_{pos}$$$, decrease $$$pos$$$ by one. If we're done with the previous step, we do the same, but while $$$pos > 1$$$ and $$$a_{pos - 1} \le a_{pos}$$$. The answer is $$$pos-1$$$.