HackerRank Staircase Problem Solution

In this post, We are going to solve HackerRank Staircase Problem.

Staircase detail:

This is a staircase of size n = 4:

   #
  ##
 ###
####

Its base and height are both equal to n. It is drawn using # symbols and spaces. The last line is not preceded by any spaces.

Write a program that prints a staircase of size n.

Function Description

Complete the staircase function in the editor below.

staircase has the following parameter(s):

int n: an integer

Print

Print a staircase as described above.

Input Format

A single integer, n, denotes the size of the staircase.

Constraints

0 < n < 100

Output Format

Print a staircase of size n using # symbols and spaces.

Note: The last line must have 0 spaces in it.

Sample Input

Sample Output

     #
    ##
   ###
  ####
 #####
######

Explanation

The staircase is right-aligned, composed of # symbols and spaces, and has a height and width of n = 6.

Table of Contents

Staircase C Solution

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    int i, k, n;

    scanf("%d", &n);

    for (i = 1; i <= n; i++) {
        for (k = 0; k < n - i; k++) {
            printf(" ");
        }

        for (k = 0; k < i; k++) {
            printf("#");
        }

        printf("\n");
    }
    
    return 0;
}

Staircase C++ Solution

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int N;
    cin >> N;
    N++;

    for (int i = 1; i < N; i++) {
        int j;
        for (j = 1; j < N - i; j++) {
            cout << ' ';
        }
        for (int k = 0; k < N - j; k++) {
            cout << '#';
        }
        cout << endl;
    }

    return 0;
}

Staircase C Sharp Solution

using System;
using System.Collections.Generic;
using System.IO;
class Solution {
    static void Main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution */
        int t = Convert.ToInt32(Console.ReadLine());
        for (int i = 1; i <= t; i++) {
            String strhash = new String('#',i);
            Console.WriteLine( strhash.PadLeft(t) );
        }        

    }

Staircase Java Solution

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    /*
     * Complete the 'staircase' function below.
     *
     * The function accepts INTEGER n as parameter.
     */

    public static void staircase(int n) {
        var nn = n;
        for(int i=1;i<=n;i++) {
            System.out.println(" ".repeat(nn-1) + "#".repeat(i));
            nn = nn - 1;
        }
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(bufferedReader.readLine().trim());

        Result.staircase(n);

        bufferedReader.close();
    }
}

Staircase JavaScript Solution

function processData(input) {
        var stair = '',
            N = +input,
            i = 1;

    while (++i < N + 2) console.log(new Array(N + 2 - i).join(' ') + new Array(i).join('#'));
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

Staircase Python Solution

size = int(input())

for i in range(1,size+1):
    print("%s%s" % (" "*(size-i), "#"*i))

Other Solutions