Sorting By Multiplication (Codeforces 1861D)
Tags: Greedy, dp

You are given an array $$$a$$$ of length $$$n$$$, consisting of positive integers. You can perform the following operation on this array any number of times (possibly zero):   choose three integers $$$l$$$, $$$r$$$ and $$$x$$$ such that $$$1 \le l \le r \le n$$$, and multiply every $$$a_i$$$ such that $$$l \le i \le r$$$ by $$$x$$$. Note that you can choose any integer as $$$x$$$, it doesn't have to be positive.You have to calculate the minimum number of operations to make the array $$$a$$$ sorted in strictly ascending order (i. e. the condition $$$a_1 < a_2 < \dots < a_n$$$ must be satisfied).

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$)  — the number of test cases.The first line of each test case contains one integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$)  — the length of the array $$$a$$$.The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$) — the array $$$a$$$.Additional constraint on the input: the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case, print one integer — the minimum number of operations required to make $$$a$$$ sorted in strictly ascending order.

Note: In the first test case, we can perform the operations as follows:   $$$l = 2$$$, $$$r = 4$$$, $$$x = 3$$$;  $$$l = 4$$$, $$$r = 4$$$, $$$x = 2$$$;  $$$l = 5$$$, $$$r = 5$$$, $$$x = 10$$$.  After these operations, the array $$$a$$$ becomes $$$[1, 3, 6, 12, 20]$$$.In the second test case, we can perform one operation as follows:   $$$l = 1$$$, $$$r = 4$$$, $$$x = -2$$$;  $$$l = 6$$$, $$$r = 6$$$, $$$x = 1337$$$.  After these operations, the array $$$a$$$ becomes $$$[-10, -8, -6, -4, 5, 1337]$$$.In the third test case, the array $$$a$$$ is already sorted.

Sample Inputs:
3
5
1 1 2 2 2
6
5 4 3 2 5 1
3
1 2 3

Sample Outputs:
3
2
0

Solution (Python):
def sol():
    n = int(input())
    a = list(map(int, input().split()))
    ans = 9999999999999999
    l = 0
    r = 0
    for i in range(n - 1):
        if a[i] >= a[i + 1]:
            l += 1
    ans = l
    a.append(9999999999)
    a.insert(0, 0)
    for i in range(1, n):
        if a[i] >= a[i - 1]:
            r += 1
        if a[i] >= a[i + 1]:
            l -= 1
        if l + r < ans:
            ans = l + r
    print(ans)

def main():
    t = int(input())
    for i in range(t):
        sol()
if __name__ == '__main__':
    main()

Solution (Python):
def main():
    t = int(input())
    r = [str(solution(i, t)) for i in range(t)]
    print('\n'.join(r))

def solution(ti, tn):
    n = int(input())
    aa = list(map(int, input().split()))
    if n < 2:
        return 0
    pref = [0] * n
    suff = [0] * n
    for i in range(1, n):
        pref[i] = pref[i - 1] + (1 if aa[i - 1] <= aa[i] else 0)
    for i in reversed(range(n - 1)):
        suff[i] = suff[i + 1] + (1 if aa[i] >= aa[i + 1] else 0)
    return min(suff[0], *(pref[i] + suff[i + 1] + 1 for i in range(n - 1)))
main()

Solution (Python):
def gao():
    ans = 0
    n = int(input())
    a = list(map(int, input().split()))
    last = [0] * n
    for i in range(n - 2, -1, -1):
        if a[i] >= a[i + 1]:
            last[i] = last[i + 1] + 1
        else:
            last[i] = last[i + 1]
    ans = last[0]
    pre = 1
    for i in range(1, n):
        ans = min(ans, pre + last[i])
        if a[i] >= a[i - 1]:
            pre += 1
    return ans
t = int(input())
ans = []
for i in range(t):
    print(gao())