just indent the code block

or just click on the "<\/>" icon in the editor.

It follows the markdown syntax: for code use 3 backticks (`) at the start and don't forget to finish it with another 3 backticks. To be sure, click Preview button.

Just that simple with JAVA StringBuilder

    public static void main(String[] args) 
    {
    Scanner in = new Scanner(System.in);
    int n = in.nextInt();

    StringBuilder builder = new StringBuilder();
    for (int i = 0; i <n ; i++) 
        builder.append(" ");
    int j = 0;

    for (int i = 1; i <=n; i++) {
        builder.replace(builder.length()-i,          
        builder.length() - j, "#");
        System.out.println(builder);
        j++;
    }

}

var cnt = Convert.ToInt32(Console.ReadLine());
var message = string.Empty;
for (int i = 1; i <= cnt; i++)
{
    Console.WriteLine("{0,"+cnt+"}", message += "#");
}

Nice!

cool!

Cheers same here N=int(raw_input())

N-i spaces i #s

for i in range(1,N+1): print ' '*(N-i)+'#'*i

can you explai it ?

(1..n).each {|a| print " "*(n-a) + "#"*a + "\n"}

For oneliners:

n=int(input());[print(" "*(n-i)+'#'*i) for i in range(1,n+1)]

wow

Excellent!

(y)

Hi, Could you Please what's wrong with this:

space=' '

n = int(input().strip())
for i in range(n):
  noOfSpaces = n - i -1
    print((space*noOfSpaces),(n-noOfSpaces)*"#")

python is powerful!

n=input()

for i in range(1,n+1): print ('#'*i).rjust(n)

python :*

A nicer way is to use rjust, I think:

for i in range(n):
    s = "#" * (i + 1)
    print(s.rjust(n, " "))

does this run asymptotically faster than c++?

I made this work in Swift by overloading the operator

var n = Int(readLine()!)!

func *(string: String, scalar:Int) -> String {
    let value = Array(count: scalar, repeatedValue: string)
    return value.joinWithSeparator("")
}

for i in 1..<n + 1 {
    print(" " * (n - i) + "#" * i)
}

https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/AdvancedOperators.html

Python's awesome for small scripts like this, but it doesn't do so well when performance matters. I tend to use it until I need fast AI/rendering. That's usually when I switch to Java or C++. Python has some of the nicest features for algorithms and data conversion.

Never knew about string multiplication, that will be useful.

n=input()
print '\n'.join(map(lambda i ('#'*i).rjust(n),range(1,n+1) ) )

cool!!

Purdy!

Python, Meet Ruby

n = gets.strip.to_i for i in 1..n do puts "#{' ' * (n-i)}#{'#' * i}" end

Great solution.

For python2 it's better to use xrange (range for python3) and I think using rjust is more pythonic.

My solution

for i in xrange(1, n+1):
    print ('#' * i).rjust(n)

Whoa! I like it.

looks so easy in python

Oh My God!

Fantastic!

Props to @deathbullet for the exact same code! xD

Wow, it's the shortest line I've seen so far

Inspired by the string arithmetic, C++ also suport such feature as followed (maybe there's a little worry for the linear space consumption, but definitely we save an extra inner loop for printing '#')

for (int i = 1; i <= n; i++){
         string s(n-i, ' ');
         string p(i,'#');
         cout << s << p << endl;
    }

but new lines ?

why + sign is used in print ? cant we do without that?

Awesome! :-)

meet meat

Here, we will use two for loops, outer for loop will print one row of star pattern ans second row will print space characters and # characters.

In any row, the sum of spaces and stars are equal to N. Number of stars increases by one and number of spaces before stars decreases by 1 in consecutive rows.

In any row R, we will first print N-R space characters then R star characters.

I did it similar

void staircase(int n) {
    // Complete this function
    int k = n;
    while (k > 0){
        for (int i = 0; i < k - 1; i++)
            printf(" ");
        for (int j = k - 1; j < n; j++)
            printf("#");
        printf("\n");
        k--;
    }
}

I created a very simple version in C:

void staircase(int n) {
    /*
     * Write your code here.
     */
    char *str= malloc(n+1);
    memset(str, (int)'#', n);
    str[n] = '\0';
    for(int i = 0; i < n; i++) {
        printf("%*.*s\n",n,i+1,str);
    }
}

nice solution! Can you explain what the underscore: "_" is doing that you pass to .map()?

You forgot to assign n in your 'oneliner'.

Note, this is technically not a single loop. Arrays.fill will iterate over the entire array to change the values.

for(int index= 0; index < arreglo.length; index++)
    arreglo[index] = ' ';