Hello coders, today we are going to solve Piling Up! Hacker Rank Solution in Python.
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
blocks = [1, 2, 3, 8, 7]
After choosing the rightmost element, 7, choose the leftmost element, 1. After than, the choices are 2 and 8. These are both larger than the top block of size 1.
blocks = [1, 2, 3, 7, 8]
Choose blocks from right to left in order to successfully stack the blocks.
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.
- 1 <= T <= 5
- 1 <= n <= 105
- 1 <= sideLength < 231
For each test case, output a single line containing either
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]
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.
Solution – Piling Up! in Python
# Enter your code here. Read input from STDIN. Print output to STDOUT ANS =  T = int(input()) for _ in range(T): n = int(input()) sl = list(map(int, input().split())) for _ in range(n-1): if sl >= sl[len(sl)-1]: a = sl sl.pop(0) elif sl < sl[len(sl)-1]: a = sl[len(sl)-1] sl.pop(len(sl)-1) else: pass if len(sl) == 1: ANS.append("Yes") if((sl > a) or (sl[len(sl)-1] > a)): ANS.append("No") break print("\n".join(ANS))
Disclaimer: The above Problem (Piling Up!) is generated by Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.