Click here to Skip to main content
15,867,568 members
Please Sign up or sign in to vote.
1.00/5 (1 vote)
See more:
Print the following pattern for a given n.
For eg. N = 6
123456
 23456
  3456
   456
    56
     6
    56
   456
  3456
 23456
123456


What I have tried:

Python
n=int(input())
i = n
while i >= 1:
    j = n
    while j > i:
        # display space
        print(' ', end='')
        j-= 1
    k = 1
    while k <= i:
        print(k, end='')
        k += 1
    print()
    i -= 1
Posted
Updated 23-Jun-22 8:21am
v2
Comments
Patrice T 18-Jul-21 10:19am    
And you have a problem ?

Python
N = 6
for i in range(1, N + 1):
    for j in range(i - 1):
        print(' ', end='')
    for j in range(i, N + 1):
        print(j, end='')
    print('')

That will do the first part, you just need to write the second part.
 
Share this answer
 
Comments
CPallini 19-Jul-21 1:58am    
5.
Richard MacCutchan 19-Jul-21 3:43am    
Thanks. I'm not sure I should admit this, but the more I use it the more I like Python.
CPallini 19-Jul-21 8:27am    
:-)
Getting your code to run at all does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
C#
int Double(int value)
   {
   return value * value;
   }

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.

Start here if you don't know how to use it: pdb — The Python Debugger — Python 3.9.6 documentation[^]

This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
 
Share this answer
 
Python


Python
<pre lang="Python"><pre lang="Python">

n = int(input( ))
for i in range(1,n+1):
    for s in range(i - 1):
        print(" ",end="")
    for j in range(i, n + 1):
        print(j, end="")
    print()
for i in range(1, n):
    for s in range(n - i - 1):
        print(" ",end="")
    for j in range(n - i, n + 1):
        print(j,end="")
    print()
 
Share this answer
 
v2

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)



CodeProject, 20 Bay Street, 11th Floor Toronto, Ontario, Canada M5J 2N8 +1 (416) 849-8900