Solving palindromic 'Triangle Quest' puzzle in Python

for i in range(1,6):
    print (((10 ** i - 1) // 9) ** 2)

Here’s a wtf one liner:

f=lambda n:n and[f(n-1),print((10**n//9)**2),range(1,n+1)];f(5)

Code golfing and taking advice of simon and rain:

set(map(lambda x:print((10**x//9)**2),range(1,N+1)))

for i in range(1, N + 1):
    print(*list(range(1, i + 1)) + list(range(i - 1, 0, -1)), sep = None)

Just because every solution offered so far involves range() which I feel is overused in Python code:

from math import log10

i = 1
while (N > log10(i)): print(i**2); i = i * 10 + 1

I am able to print in list format using below:

    for i in range(1,5):
        print [j for j in range(1,i+1) ], [j for j in range(i-1,0,-1) ]

Result:

[1] []
[1, 2] [1]
[1, 2, 3] [2, 1]
[1, 2, 3, 4] [3, 2, 1]
[1, 2, 3, 4, 5] [4, 3, 2, 1]

def palindrome(N):
    for i in range(1, N + 1):
        print(int('1' * i)**2)

palindrome(int(input()))

• 1 * 1 = 1
• 11 * 11 = 121
• 111 * 111 = 12321

for i in range(1,int(input())+1):
   print(int((10**i-1)/9)**2)

1 -> (   10 - 1) / 9 =    1,    1 *    1 = 1
2 -> (  100 - 1) / 9 =   11,   11 *   11 = 121
3 -> ( 1000 - 1) / 9 =  111,  111 *  111 = 12321
4 -> (10000 - 1) / 9 = 1111, 1111 * 1111 = 1234321

    for i in range(2,int(raw_input())+2): 
        print ''.join(([unicode(k) for k in range(1,i)]))+""+''.join(([unicode(k) for k in range(i-2,0,-1)]))
        print ''.join(map(unicode,range(1,i)))+""+''.join(map(unicode,range(i-2,0,-1)))

I hope it will help.

I think the following code should work. I have used the most basic method, so that most people are going to understand it:

  N = int(input())
  arr = []
  for i in range(1,N+1):
         arr.append(i)
         print(arr+arr[-2: :-1])

Use this code:

prefix = ''
suffix = ''

for i in range(1,n):
    middle = str(i)
    string = prefix + middle + suffix
    print(string)

    prefix = prefix + str(i)
    suffix = ''.join(reversed(prefix)) 

for i in range(1,int(input())+1):
    print(int(str('1'*i))**2)

Basically on each iteration, the number ‘1’ in string is multiplied by i times then converted to an integer then squared. So e.g.

On the 3rd iteration –> output = 111^2 = 12321

Edit: Noticed constraints was to answer with str() function

So we have a sequence 1, 11, 111, 1111
nth = an + (a(r^(n-1) – 1)) / (r – 1) where |r > 1|

therefore, solution;

for i in range(1,int(input())+1):
    print(pow((((10**i - 10))//9) + 1, 2))

  for i in range(1,int(input())+1): #More than 2 lines will result in 0 score. Do not leave a blank line also
    print(''.join(list(map(lambda x:str(x),list(range(i+1))[1:]))+list(map(lambda x:str(x),list(reversed(list(range(i))[1:]))))))

This is a simple string-less version:

for i in range(1, int(input()) + 1):
    print(sum(list(map(lambda x: 10 ** x, range(i)))) ** 2)

I know this has been asked a while ago, but I just stumbled upon this exercise and found an alternative and elegant solution to it.

As suggested by @raina77ow, we know that 11 * 11 = 121, 111 * 111 = 12321 and so on.
But we also know that:

2**1 - 1 = 1 (is 1 in binary)
2**2 - 1 = 3 (is 11 in binary)
2**3 - 1 = 7 (is 111 in binary)
2**4 - 1 = 15 (is 1111 in binary)
etc.

Now, by doing bin(15) you get 0b1111 and I hate that I had to transform it to integer using slicing, but I found no other method.

So, using the above, this is my solution:

for i in range(1,int(input())+1):  # given line
    print(int(bin(2**i - 1)[2:])**2)

Putting in my two cents worth, using unpacking of range

    for i in range(1, int(input()) + 1):
        print (*range(1, i+1), *range(i-1, 0, -1))

Just noticed Keith Hall has a very similar solution. He should get original credit for this.