I can't comment in formatted way. Can you tell me how to do that?

I was looking for exactly that type of string method. I figured c# just had to have something like that. Thanks fer sharin'.

It needs less coding, but it is not a fast approach. This String constrcutor overload is too slow.

Check here: http://www.codeexplain.com/code/0cecd228-c32e-4a5a-b136-2461c4743a99 "I can get Substring to be 5-6 times faster but using a constructor like String(Char, Int32) turns out to be atleast 3 times slower"

That is ugly. Meet Python

n = int(raw_input())
for i in range(1,n+1):
    print " "*(n-i) + "#"*i

Wow. Smart way to code. Thanks for sharing this!!!

you actually can remove the second parameter in the padleft method

Even easier in python

n = int(raw_input().strip())
for i in reversed(xrange(n)):
    print (' ' * (i)) + ('#' * (n - i))

Thank you so much, but would you mind telling me why use PadLeft(N,' ') instead of PadLeft(N-1,' ')? The 1st N=6 right? I tried to use N-1, but it only prints out 4 spaces on the first line.

I didn't know about the PadLeft option. 'Preciate your sharing this.

Also, you can start the iterator at i=1 to avoid the "i+1" in the string constructor, then run it until i<=N. It's all the same, of course, but a little more intuitive for folks like me.

Easy Ruby:

(1..n).step(1) { |i| str = " "*(n-i)+("#"*i); puts str }

I used a slightly different approach:

int size = Convert.ToInt32(Console.ReadLine());
        
        for(int i=1;i<=size;i++){            
            Console.WriteLine(new string(' ', size - i) + new string('#', i));            
        }

Interesting. I used Enumerable.Repeat and concatenating a string. I prefer your padleft approach though

int n = Convert.ToInt32(Console.ReadLine());
for(int i = 0; i < n; i++)        
{            
    string w = string.Join("", Enumerable.Repeat(" ", (n-1) - i));
    w += string.Join("", Enumerable.Repeat("#", i + 1));      
    Console.WriteLine(w);
}

I will simplify it like this:-

for(int i=1;i <= N ; i++) Console.WriteLine(new String('#',i).PadLeft(N));

Did a similar thing

for(int i = 1; i < n+1; i++) {
    Console.WriteLine(new string (' ', (n-i)) + new string ('#', (i)));
}

I did not know that String overload :/ My solution:

using System;
class Solution
{
    static void Main(String[] args)
    {
        int N = int.Parse(Console.ReadLine());
        for (int i = 0; i < N; i++)
            Console.WriteLine(Clone('#', i).PadLeft(N, ' '));
    }
    
    private static string Clone(char character, int count)
    {
    	return "".PadLeft(count, character);
    }
}

Did it like this in C. Could this get shorter?

int main(){
    int n;
    scanf("%d",&n);
    int a = n;
    char space = ' ';
    char hash = '#';
    
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < a - 1; j++)
        {
            printf("%c", space);
        }
        
        a--;
        
        for(int k = 0; k < i + 1; k++)
        {
            printf("%c", hash);
        }
        printf("\n");
    }
    return 0;
}

Its too easy with Python! :) here..

n=int(input())
for i in range(1,n+1):  print(' '*(n-i)+'#'*i)

more easy in python one line

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

include

using namespace std;

int main() { int n; cin >> n; for(int i=0;in-i-2) cout<<"#"; else cout<<" "; } cout<

JavaScript one-liner (not the most efficient :P)

const staircase = n => console.log(new Array(n).fill(0).map((_, idx) => "#".repeat(idx + 1).padStart(n)).join('\n'));

One of the best and easy solution

Thanks

In one line: [print(' '*(n-x) + '#'*(x)) for x in range(1,n+1)]

I like this solution. Clean and easy. I did two prints and a custom range but this good. =)

I really like this solution. Only needs a single loop and string!

I wrote a solution using java primitives... which is what most interviewers are looking for. Java collections can be tempting though.

I did this:

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

public class Solution {

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

        for (int i = 1; i <= n; i = i + 1) {
            int pt = n - i;
            for (int c = 1; c <= pt; c = c + 1) {
                System.out.print(" ");
            }
            for (int c = 1; c <= i; c = c + 1) {
                System.out.print("#");
            }
            if (i < n) {
                System.out.println();
            }
        }
    }
}

Another approach is:

    StringBuilder sb = new StringBuilder(n);
    String fmt = "%"+n+"s\n";
    for (int i=1;i<=n;i++){
        sb.append('#');
        System.out.printf(fmt, sb);
    }

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

    for(int i = 1 ; i <= n ; i ++){
        String ws = " ";
        String hash = "#";
        String repeatWs;
        String repeatHash;
        repeatWs = new String(new char[n-i]).replace("\0",ws);
        repeatHash = new String(new char[i]).replace("\0",hash);
        System.out.println(repeatWs+repeatHash);
    }
}

Can you shoew me how do it in Javascript?

This is mine:

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

    for(int i = 1; i <= n; i++){
        String padded = "";
        padded = String.format("%-" + i + "s","#").replace(' ','#');
        padded = String.format("%" + (n) + "s",padded);
        System.out.println(padded);
    }
}

nyc work ...it helped me

Excelent Solution!!! That's quite simple to read and it works fine.

Congratulations!!!!

This is my solution for java8

public static void staircase(int n) {
		int s = n-1;
		int h = 1;
		while (n != 0) {
			for (int i = 0; i < s; i++) {
				System.out.print(" ");
			}
			for (int j = 0; j < h; j++) {
				System.out.print("#");
			}
			System.out.println();
			h++;
			s--;
			n--;
		}
	}

This is by far the best solution I've seen on this thread. Easy to understand.

Just did a similar solution in C++. I must to be quite efficient

inline void patch_buffer(std::string & buffer, int i) {
        // replace the i-th character of a buffer with a dash
        buffer[i] = '#';
}

void staircase(int n) {
        // fill the buffer with spaces
        std::string buffer = std::string(n, ' ');
        const int max_index = n - 1;

        for(int i = max_index; i >= 0; i--) {
                // for each line the (max_index - i)-th character of
                // a buffer will be replaced with a dash
                patch_buffer(buffer, i);

                // and then the whole line will be printed
                std::cout << buffer << "\n";
        }
}

People that want to try something special may try to write an iterator with one range-based printing loop.