human
Can you provide a Python code to find the minimal possible number of inversions in the deque after the whole array is processed given the following:
Can you solve the below in PYTHON3?
In fact, the problems E1 and E2 do not have much in common. You should probably think of them as two separate problems.

You are given an integer array a[1 … n] = [a_1, a_2, …, a_n].

Let us consider an empty [deque](https://tinyurl.com/pfeucbux) (double-ended queue). A deque is a data structure that supports adding elements to both the beginning and the end. So, if there are elements [3, 4, 4] currently in the deque, adding an element 1 to the beginning will produce the sequence [\color{red}{1}, 3, 4, 4], and adding the same element to the end will produce [3, 4, 4, \color{red}{1}].

The elements of the array are sequentially added to the initially empty deque, starting with a_1 and finishing with a_n. Before adding each element to the deque, you may choose whether to add it to the beginning or to the end.

For example, if we consider an array a = [3, 7, 5, 5], one of the possible sequences of actions looks like this:    1. |  add 3 to the beginning of the deque: |  deque has a sequence [\color{red}{3}] in it.
  
---|---|---  
  2. |  add 7 to the end of the deque: |  deque has a sequence [3, \color{red}{7}] in it.
  
  3. |  add 5 to the end of the deque: |  deque has a sequence [3, 7, \color{red}{5}] in it.
  
  4. |  add 5 to the beginning of the deque: |  deque has a sequence [\color{red}{5}, 3, 7, 5] in it.
  
  
Find the minimal possible number of inversions in the deque after the whole array is processed. 

An inversion in sequence d is a pair of indices (i, j) such that i < j and d_i > d_j. For example, the array d = [5, 3, 7, 5] has exactly two inversions — (1, 2) and (3, 4), since d_1 = 5 > 3 = d_2 and d_3 = 7 > 5 = d_4.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases.

The next 2t lines contain descriptions of the test cases. 

The first line of each test case description contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — array size. The second line of the description contains n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array.

It is...
assistant
#Here is Python code to find the minimal possible number of inversions in the deque after the whole array is processed.

def naiveSolve():
    
    
    
    return


def solve():
    
    
    
    return

class SegmentTree():
    def __init__(self,arr,combinerFunction,isCommutative=True): # isCommutative means f(a,b)==f(b,a)
        self.n=len(arr)
        def cmb(a,b):
            if a==None: return b
            if b==None: return a
            return combinerFunction(a,b)
        self.cmb=cmb
        self.sz=self.n if isCommutative else pow(2,((self.n-1).bit_length())) # non-commutative needs power of 2 size
        self.t=[None]*(2*self.sz) # use None as initial value
        for i in range(self.n): self.t[i+self.sz]=arr[i]
        for i in range(self.sz-1,0,-1): self.pull(i)
    def pull(self,p):
        self.t[p]=self.cmb(self.t[2*p],self.t[2*p+1])
    def update(self,idx,val): # set val at idx
        idx+=self.sz
        self.t[idx]=val
        idx//=2
        while idx>0:
            self.pull(idx)
            idx//=2
    def query(self,l,r): # aggregated value in [l,r] inclusive
        l+=self.sz; r+=self.sz+1
        a=b=None
        while l<r:
            if l%2==1: a=self.cmb(a,self.t[l]); l+=1
            if r%2==1: r-=1; b=self.cmb(self.t[r],b)
            l//=2; r//=2
        return self.cmb(a,b)
def getMaxSegTree(arr):
    return SegmentTree(arr,lambda a,b:max(a,b),True)
def getMinSegTree(arr):
    return SegmentTree(arr,lambda a,b:min(a,b),True)
def getSumSegTree(arr):
    return SegmentTree(arr,lambda a,b:a+b,True)

def main():
    
    # notice that for each number added, the order of the prefix does not
    # its number of inversions. hence, just add to the side that creates
    # fewer inversions. use seg tree and coord compression
    
    t=int(input())
    allans=[]
    for _ in range(t):
        n=int(input())
        a=readIntArr()
        
        c=sorted(set(a))
        reverseMap={x:i for i,x in enumerate(c)}
        for i in range(n):
            a[i]=reverseMap[a[i]]
        
        maxN=max(a)+2
        st=getSumSegTree([0]*maxN)
        cnts=[0]*maxN
        ans=0
        for x in a:
   ...