nice puzzle mate but it seems pretty confusing

brute force

Is this code working??

if input is 0 , I think it shouldn't print any row .

Same as my way, but in JavaScript.

fake

Exactly what I came up with!

How about this?

for i in 1...n {
    var string = ""
    string = String(repeating: " ", count: n-i)
    string.append(String(repeating: "#", count: i))
    print(string)
}

good to see your implemenation , by seeing your code i found that i was uncessary using an extra loop

can you please explain how this line wokrs

System.out.printf("%15s","##"+"\n");

A small change in your code to make it more simple

public class Solution {

public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();
    String str="#";
    for (int i=0;i<n;i++)
        { 
        System.out.printf("%"+n+"s%n",str);
        str=str+"#";
    }
}

}

great!! gud job

Why is it n+1. Isn't that saying how long the string is so it should be n for all?

System.out.printf("%"+(n+1)+"s",str+"\n");

hey i am new to java can u pls explain ur code

What's the meaning of % and s in printf statement? Please help.

please can you explain this line System.out.printf("%"+(n+1)+"s",str+"\n");

Beautiful! I like the simplicity of your solution :)

can you explain the line ("%"+(n+1)+"s",str+"\n")

in your " System.out.printf("%"+(n+1)+"s",str+"\n");" i am getting confuse.,str+"\n"-it will give # but "%"+(n+1)+"s" ?..

Awsome...but how is it working...can someone please exaplin details..

Much better than my for-loop ideas! Thanks for sharing.

can you please explain this line i didn't got that

System.out.printf("%"+(n+1)+"s",str+"\n");

Just in love with your precise code!!!! Lit

Sorry Anoop, i may not help you, i am expert in python notin other languages....

Here is the video explanation –

https://youtu.be/gDltV7pJw7A

My answer was similar, I concatenated instead of justified

[print(' '*(n-i) + '#'*i) for i in range(1,n+1)]

if you don't want to use rjust. You can simply add space yourself

for length in range(n):
    print(' '*(n-length-1)+'#'*(length+1))

one liner

(lambda num: [print(('#'*x).rjust(num)) for x in range(num+1)])(int(input().strip()))

for pound, space in zip(list(range(1,n+1)), reversed(list(range(1,n+1)))):
                print(' '*space + '#'*pound)

Is there something wrong with this answer? Cannot pass the test case.

rjust does not add n spaces to the string, it makes the string n spaces long in total

Consider the example "john".rjust(5) This doesnt actually give 5 spaces to the string. Consider the given string as a block [j][o][h][n](i.e there are four blocks) As we have given width=5 and the given string is of 4 blocks ,it just adds one more block(i.e,the empty block which is a space) to make it equal to 5. [][j][o][h][n] So now it becomes " john".Make note of the space before the string. If you specify any char,the space then gets filled with the given char.

firstly, he prints it without the spaces, so it looks like this

#
##
###
####

then he uses rjust just to right justify the above string and fill the empty gaps with spaces. Afer that, it looks like this:

   #
  ##
 ###
####

Sorry for my bad english

try this

count = 1
while(count <= n):
    print(('#' * count).rjust(n))
    count += 1

Thanks :)

pls explain this code,cant understand.

I did it a little less concisely:

line = 1
stair = ""
while line <= n:
    stair += "#"
    print(str(stair).rjust(n))
    line += 1

show me your code?

simpler one count = 1 while(count <= n): print(('#' * count).rjust(n)) count += 1

Realized that I never put the answer to my own question here in case someone else was curious about the same thing.

This technique is covered in the Python documentation. Although, it's not covered in the Print Methods or String section of the documentation, it is in the Built-in Types Section.

https://docs.python.org/3/library/stdtypes.html#typesseq-common

Have been reading this section and learning a lot; hopefully it will be as useful for the rest of the programming enthusiasts out there!