Thanks BlueBird this really helped me a lot

how the builder.length() - j works at first time?

You really don't need more than one loop for this and can do with formatted output

this code also works well. time complexity is O(N).

public static String repeat(String str, int times) {
    return new String(new char[times]).replace("\0", str);
}

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

    for(int i = n-1; i >= 0; i--) {
        System.out.print(repeat(" ", i));
        System.out.println(repeat("#", n-i));
    }
}

nice method

O(n) algorithm here. Two nested for loops are O(n^2) algorithm.

I did it like this ...

import java.io.*;
import java.util.*;

public class Solution {
    static String getFloor(int size, int i) {
        String strFloor = "";
        int n = size - (i + 2);
        for(int k = 0; k < size; ++k) {
            if(k > n)
                strFloor += "#";
            else
                strFloor += " ";            
        }
        return strFloor;
    }    
    
    public static void main(String[] args) {        
        Scanner sc = new Scanner(System.in);
        int size = sc.nextInt();
        
        for(int i = 0; i < size; ++i) {            
            System.out.println(getFloor(size, i));
        }
    }
}

This problem reminds me with a similar problem that I faced during the early semesters of my Bachelor study :) At that time, I used the inefficient nested loop solutions. Turns out, there's a way better solution posted here. Thanks!

Here's my solution (inspired by your solution):

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static void staircase(int n) {
        // Complete this function
        StringBuilder sb = new StringBuilder();
    
        //create the spaces
        for(int i = 0;i<n-1;i++){
            sb.append(" ");
        }
        
        //create the pound signs (replace the spaces to the pound sign)
        for(int j=n;j>0;j--){
            sb.replace(j-1, j, "#");
            System.out.println(sb);
        }  
    }
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        staircase(n);
        in.close();
    }
}

This is interesting. I've tried to look up "{0, "+ +"}" to find out more about it, but with no luck. Is there name for this formatting type? I'm curious about this specific part: "+cnt+"

for i in range(1,n+1)

sets i to 1 then 2 then 3 then 4 then 5 then 6 then 7

string multiplication simply repeats the string i times;

3 * "A" = "AAA"

yes i ca explai

I understand the words "print" and "each." Couldn't you get rid of those bits?

(1..n).each { |i| puts "#{'#' * i}".rjust(n)}

That seems much easier.

Or for those who feel a semicolon in a one-liner is cheating, Python2:

print '\n'.join(('#' * (i + 1)).rjust(n) for n in [int(raw_input())] for i in xrange(n))

Python3:

[print(('#' * (i + 1)).rjust(n)) for n in (int(input()),) for i in range(n)]

for java one liner's

while(++i <= n)
System.out.println(new String(new char[n-i]).replace("\0", " ")+new String(new char[i]).replace("\0", "#"));  

I don't have Python installed so I can't run it. What is the output you are getting or error message? Why is the "print" statement indented at the same level as "noOfSpaces"?

This is pretty elegant. And I've created a monster:

var array = Array(arrayLiteral: space)
for i in 1...n-1{
    array.append(" ")
}
array[n - 1] = "#"
for i in 1...n{
    array[n - i] = "#"
    var string = array.joinWithSeparator("")
    print(string)
}

Great! my way is trying to stitch sring directly via '+', silly.

My solution:

let space = Character(" ")
let char = Character("#")

var spaceCount = 0
var charCount = 0

for (var i = 0; i<n; i++) {
    spaceCount = n-1-i
    charCount = i+1
    let spaceString = String(count: spaceCount, repeatedValue: space)
    let charString = String(count: charCount, repeatedValue: char)
    print("\(spaceString)\(charString)")
}

how about this ?

for i in 1...n {
    print(String(repeating: " ", count: n - i) + String(repeating: "#", count: i))
}

I used string concatenation in my code's print statement, but I like this better! I'd never had cause to use rjust before, so it didn't even occur to me. This solution has the added bonus of actually following the problem description (i.e., "The staircase is right-aligned...").

Now I've learned something new. Cheers to you and nicolewhite and others who used rjust!

Please can someone explain me the code ,i have never used this kind of definition of string . And could someone please tell if there is any method to do it using setw() ,setfill() and stuff ,although i think setfill() wouldnt be required .

amazzzzzzzzzzzzzzzzzzzzzzing ..

Can there be any other optimized way other than above ?

in.close();

can you explain your code ?

for (var i = 1; i <= n; i++) {
    console.log(' '.repeat(n - i) + '#'.repeat(i));
}

I took a similar approach using arrays

var padding = new Array(n + 1).join(' ');
var hashes = new Array(n + 1).join('#');
for(var i=1;i<=n;i++){
    console.log(padding.slice(i) + hashes.slice(n-i));
}

Sure.

Basically this expression (%-" + i + "s","#") means:

Apply this format (%-" + i + "s") to the parametrized String #. 

%: means a formatting expression will follow.
-: means you want to padd to the rigth.( + or none means that you want to padd to the left).
i: means the length of the string that you expect. The length of the missing characters will filled with blank spaces.
s: denotes the type of the object, in this case strings.

For each iteration I get:

'#' where i=1. No padding is added because you expect a string of length 1.
'# ' where i=2.
'#  ' where i=3.
.... until '#      ' where i = 6.

For each iteration I replace blank spaces with # (.replace(' ','#');). So finally I get:

'#'
'##'
'###'
'####'
'#####'
'######'

For each iteration also I apply a final formatting:

String.format("%" + (n) + "s",padded).

Basically I am saying I expect in every iteration a string with a length of 6 characteres and every missing position will filled with blank spaces to the left. So I get:

'     #'
'    ##'
'   ###'
'  ####'
' #####'
'######'

And that's all.

could you explain what the code is doing there?

this is magic to me.

Good one!! :)

@meaux, that code takes advantage of stream manipulators to format the output buffer: with setw(n) you define the width of the field to display, setfill(' ') lets you choose a padding character, and string(i,'#') builds a string consisting of i times the '#' character.

I had also gone for a one liner inside the loop (not really, as I had to reset a variable on every loop iteration) using std::transform and lambdas on a char array, but this is way more elegant and readable.

Java

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