HackerRank Piling Up! problem solution in Python

HackerRank Piling Up! problem solution in Python – In this Piling Up! problem, we need to develop a Python program that can read a single integer containing the number of test cases. And then we need to print yes or no on the output screen.

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 cub[i] is on top of cub[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.

HackerRank Piling Up! problem solution in Python 2.

# Enter your code here. Read input from STDIN. Print output to STDOUT
tests = int(raw_input().strip())
while tests > 0:
  n = int(raw_input().strip())
  arr = map(int, raw_input().strip().split(' '))
  Lmin = arr[:]
  Rmin = arr[:]
  for i in range(1,n):
    Lmin[i] = min(Lmin[i-1], arr[i])
  for i in range(n-2, -1, -1):
    Rmin[i] = min(Rmin[i+1], arr[i])
  result = 'Yes'
  for i in range(1,n-1):
    if Lmin[i] < arr[i] and Rmin[i] < arr[i]:
      result = 'No'
      break
  print result
  tests -= 1

Piling Up! Problem solution in Python 3.

# Enter your code here. Read input from STDIN. Print output to STDOUT
for t in range(int(input())):
    input()
    lst = [int(i) for i in input().split()]
    min_list = lst.index(min(lst))
    left = lst[:min_list]
    right = lst[min_list+1:]
    if left == sorted(left,reverse=True) and right == sorted(right):
        print("Yes")
    else:
        print("No")

Problem solution in pypy programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT
from collections import deque
n = int(raw_input())

for _ in range(n):
    m = int(raw_input())
    cubes = deque(map(int, raw_input().strip().split()))
    cube = ''
    for _ in range(m):
        left, right = cubes[0], cubes[-1]
        if left >= right:
            if cube >= left or cube == '':
                cube = left
                cubes.popleft()
        else:
            if cube >= right or cube == '':
                cube = right
                cubes.pop()
            
    if len(cubes) == 0:
        print 'Yes'
    else:
        print 'No'
    

Problem solution in pypy3 programming.

# https://www.hackerrank.com/challenges/piling-up

if __name__ == '__main__':

    for i in range(int(input())):
        num = int(input())
        last_num = 99999999999
        arr = map(int, input().split())
        items = list(arr) 
        
        start = 0
        end = len(items) - 1
        while start <= end:
            if items[start] > items[end]:
                if items[start] <= last_num:
                    last_num = items[start]
                    # print(items[start], last_num - items[start])
                    start += 1
                    
                else: 
                    print('No')
                    break
                
            else:
                if items[end] <= last_num:
                    last_num = items[end]
                    # print(items[end], last_num - items[end])
                    end -= 1
                    
                else: 
                    # print(last_num - items[end])
                    print('No')
                    break
                
        if start > end:
            print('Yes')
        # else: print('NO')