1. Sort cubes list as an expected stack to be find
2. Check if picking only left or right cubes is possible to pile the expected stack

T = int(input())

def is_stackable(cubes):
    expected_stack = sorted(cubes)
    
    left, right = 0, len(cubes) - 1
    while len(expected_stack) > 0:
        expected_block = expected_stack.pop()
        
        if cubes[left] == expected_block:
            left += 1
        elif cubes[right] == expected_block:
            right -= 1
        else:
            return False
        
    return True

for t in range(T):
    n = int(input())
    cubes = list(map(int, input().split()))
    print("Yes" if is_stackable(cubes) else "No")

'''if the largest block isn't in the first or last position to begin with, the blocks are not stackable'''

for _ in range(int(input())): n = int(input()) blocks = list(map(int, input().split())) if blocks[0] == max(blocks) or blocks[-1] == max(blocks): print("Yes") else: print("No")

This is a solution without any dependencies

for i in range(int(input())):
    l=int(input())
    left,right=0,-1
    ls=list(map(int,input().split()))
    prev=0
    for k in range(l):
        ind=0
        if ls[left]>ls[right]:
            ind=left
        elif ls[right]>ls[left]:
            ind=right
        else:
            ind=left
        if(ls[ind]<=prev) or (prev==0):
            prev=ls[ind]
            ls.pop(ind)     
    if ls==[]:
        print("Yes")
    else:
        print("No")

1. n = int(input())
2. for i in range(n):
3. len_blocks = input()
4. block = list(map(int,input().split()))
5. li = []
6. l,r = 0 , len(block)-1
7. while(l<=r):
8. if(block[r]>=block[l]):
9. li.append(block[r])
10. r -=1
11. else:
12. li.append(block[l])
13. l+=1
14. if((li == sorted(li, reverse=True)) == True):
15. print("Yes")
16. else:
17. print("No") '''

t = int(input()) for _ in range(t): n = int(input()) ls = list(map(int, input().split())) k = float('inf')

left, right = 0, n - 1
while left <= right:
    if k < max(ls[left], ls[right]):
        print("No")
        break
    if ls[left] > ls[right]:
        k = ls[left]
        left += 1
    else:
        k = ls[right]
        right -= 1
else:
    print("Yes")