HackerRank "Larry's Array"

I caught the sparkle in my mind and got AC1 ! It is a great great experience !

So the basic idea: permute on 3 consecutive items doesn't change the parity of the no. of inversions. Each permutation can only remove 0 or 2 inversions. So we say "YES" when no. of inversion % 2 = = 0. And we use MergeSort to count it in O(nlgn).

ret = 0
def merge(arr1, arr2):
    global ret
    if not arr1: return arr2
    if not arr2: return arr1
    for v2 in arr2:
        arr1.append(v2)
        i = len(arr1) - 1
        while(i > 0):
            if arr1[i] < arr1[i - 1]:
                arr1[i - 1],arr1[i] = arr1[i], arr1[i - 1]
                i -= 1
                ret += 1
            else:
                break
    return arr1

def mergeSort(arr):
    n = len(arr)
    if (n < 2): return arr
    return merge(mergeSort(arr[:n//2]), mergeSort(arr[n//2:]))

###
t = int(input())
for _ in range(t):
    n = int(input())
    arr = list(map(int, input().split()))
    ret = 0
    mergeSort(arr)
    print("YES" if ret % 2 == 0 else "NO")