## Algorithm

Problem Name: Python - Piling Up!

In this HackerRank Functions in PYTHON problem solution,

There is a horizontal row of n cubes. The length of each cube is given. You need to create a new vertical pile of cubes. The new pile should follow these directions: if cube[i] is on top of cube[j] then sideLength[j] >= sideLength[i]

When stacking the cubes, you can only pick up either the leftmost or the rightmost cube each time. Print Yes if it is possible to stack the cubes. Otherwise, print No.

Example

blocks = [1,2,3,8,7].

Result: Yes

Choose blocks from right to left in order to successfully stack the blocks.

Input Format

The first line contains a single integer T,  the number of test cases.
For each test case, there are 2 lines.
The first line of each test case contains n the number of cubes.
The second line contains n space separated integers, denoting the sideLengths of each cube in that order.

Constraints

1 <= T <= 5

1 <= n <= 10**5

1 <= sideLength <= 2**31

Output Format

For each test case, output a single line containing either Yes or No.

Sample Input

STDIN        Function
-----        --------
2            T = 2
6            blocks[] size n = 6
4 3 2 1 3 4  blocks = [4, 3, 2, 1, 3, 4]
3            blocks[] size n = 3
1 3 2        blocks = [1, 3, 2]


Sample Output

Yes
No


Explanation

In the first test case, pick in this order: left -4 , right -4, left -3, right -3, left -2, right -1.

In the second test case, no order gives an appropriate arrangement of vertical cubes. 3 will always come after either 1 or 2.

## Code Examples

### #1 Code Example with Python Programming

Code - Python Programming


from collections import deque

for _ in range(int(input())):
input()
dq = deque(map(int, input().split()))
last = None
while dq:
curr = dq.popleft() if dq[-1] < dq else dq.pop()
if last and last < curr:
break
else:
last = curr

print('No' if dq else 'Yes')

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