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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++;
}
}
publicstaticvoidmain(String[]args){Scannerin=newScanner(System.in);intn=in.nextInt();in.close();//char c = '#';for(inti=0;i<n;i++){for(intj=0;j<=n-i-2;j++){System.out.print(" ");}for(intj=n-i-1;j<n;j++){System.out.print("#");}System.out.println();}}
functioncreateString(str,ns,n){if(ns>0)returncreateString(str+" ",ns-1,n);elseif(n>0&&ns===0)returncreateString(str+"#",ns,n-1);elseif(n===0&&ns===0)returnstr;}// Complete the staircase function below.functionstaircase(n){letns=n.length-1;console.log(createString("",ns,n));}
i was going to add a while loop but i keep getting undefined not sure the problem.
And then I print a number of # amounting to the current loop index on each line followed with a newline break.
But since this will lead to the output being aligned on the left,
I also print (n-i) spaces on each line where n is the maximum number and i is the current loop index before executing (2). So for instance when i is 1, I print 5 spaces before I print 1 # to fill up the six spaces (total number of spaces on each line). When i is 2, I print 4 spaces before I print 2 # to make sure all the six spaces on the line are filled.
This also ensures that the # is aligned to the right.
NB. The range is (1, n+1). I didn't start from 0 because that will cause an empty space at the beginning of the output and I also ended at n+1 because the loop ends and n-1 so I add 1 to get to the correct n.
You can't get better than O(n^2), the goal is to print an nxn matrix where some of the elements are spaces and some are #. In order to do that you have to do at least the n^2 print statements.
Just because you can code it in a few lines in a language like C# doesn't mean it's more efficient, that same process is going on behind the scenes.
The printf is for formatting and when we give %s it indicate the String need to format, the number indicate (15) starting point for printing string (##) and \n indicate for new line so that ## will start printing in a new Line and start from 15th space
for more details refer
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+"#";
}
}
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.
Could you please format your code or do one thing if its not in java then cross verify your solution with mine one. It will help you to figure out the mistake at your own.
Good Evening! I was amazed at your succinct and easy solution to this problem. I am currently getting my feet wet in Python and have been using Hackerrank in order to improve my familiarity with it along with its online documentation.
I main question is this: How did you know that you could use
print('#'*1)
I have scoured programiz.com and Stack Overflow for ideas on handling this problem, but this technique never came up.
Any lead on becoming a better python programmer would be greatly appreciated!
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.
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));
}
}
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):
importjava.io.*;importjava.util.*;importjava.text.*;importjava.math.*;importjava.util.regex.*;publicclassSolution{staticvoidstaircase(intn){// Complete this functionStringBuildersb=newStringBuilder();//create the spacesfor(inti=0;i<n-1;i++){sb.append(" ");}//create the pound signs (replace the spaces to the pound sign)for(intj=n;j>0;j--){sb.replace(j-1,j,"#");System.out.println(sb);}}publicstaticvoidmain(String[]args){Scannerin=newScanner(System.in);intn=in.nextInt();staircase(n);in.close();}}
Staircase
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I was looking for exactly that type of string method. I figured c# just had to have something like that. Thanks fer sharin'.
Just that simple with JAVA StringBuilder
Hows it?
I had the same general idea, but it's not passing even though the output is identical :/
edit: crap. changed the CR to newline and it passed.
My idea
nice puzzle mate but it seems pretty confusing
How about this mate
might
3 nested for loops? There's def. something to be improved here, think about recursion.
its really only 2 nested for loops and therye each operating on different ranges.
2 nested loops and the size is extremely small so computation time is not something we need to consider
how do you see my Python 2 solution?
i was going to add a while loop but i keep getting undefined not sure the problem.
okay im an idiot and for some reason put n.length as if a int has a length.
fantastic idea dude
can you explain this ?
That was my solution basically as well. I like it ;)
i m still confused with the logic, can you please explain it furthure more?
NB. The range is (1, n+1). I didn't start from 0 because that will cause an empty space at the beginning of the output and I also ended at n+1 because the loop ends and n-1 so I add 1 to get to the correct n.
Do you get it now?
Yes! you explained it awesome. Thank you man!
void staircase(int n) { int i,j; for(i=n;i>0;i--) { for(j=1;j<=n;j++) { if(j>=i) { printf("#"); } else printf(" ");
}
} thankx man
brute force
What?! Because it's not convoluted crap it's brute force? Only silly thing is that he started indices at 0 instead of 1.
It's not the most efficient at O(n^2)
You can't get better than O(n^2), the goal is to print an nxn matrix where some of the elements are spaces and some are #. In order to do that you have to do at least the n^2 print statements.
Just because you can code it in a few lines in a language like C# doesn't mean it's more efficient, that same process is going on behind the scenes.
yes,bro it you are saying correct...it will minimum require of O(n^2) time complexity
static void staircase(int n) {
string str = ""; for (int i = 0; i < n; i++) { str += "#"; Console.Write(str.PadLeft(n,' ')+"\n"); }
agreed - no need for 2 nested loops
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?
similar to my Scala solution. What programming language is this?
Python2
I was referring to zeel_mehta97 code. Somehow I fail to see Python there.
for i in 1...n { var string = "" string = String(repeating: " ", count: n-i) string.append(String(repeating: "#", count: i)) print(string) }
I inspired by your solution.
good to see your implemenation , by seeing your code i found that i was uncessary using an extra loop
The "extra" loop is hidden by the printf. So what? Complexity is the same.
can you please explain how this line wokrs
System.out.printf("%15s","##"+"\n");
The printf is for formatting and when we give %s it indicate the String need to format, the number indicate (15) starting point for printing string (##) and \n indicate for new line so that ## will start printing in a new Line and start from 15th space for more details refer
*https://docs.oracle.com/javase/tutorial/java/data/numberformat.html *
it starts counting from 0 or 1 for the places
A small change in your code to make it more simple
public class Solution {
}
Use a StringBuilder instead of String
PLEASE EXPLAIN IT
great!! gud job
Why is it n+1. Isn't that saying how long the string is so it should be n for all?
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
Hello Mansi,
I think this video can help you to understand the above mentioned line.
Here is the video explanation –
https://youtu.be/gDltV7pJw7A
Just in love with your precise code!!!! Lit
Can you explain me how ur code works
Safer to use Environment.NewLine
how about this?
String hash = "#"; for(int i = 1; i <= n; i++){ System.out.printf("%"+n+"."+i+"s%n", hash); hash+="#"; }
Sorry Anoop, i may not help you, i am expert in python notin other languages....
Can you please explain
Here is the video explanation –
https://youtu.be/gDltV7pJw7A
python is love,,,,python is life
nice,
AMAZING
I updated my answer based on yours
My answer was similar, I concatenated instead of justified
like my Scala solution, same idea.
if you don't want to use rjust. You can simply add space yourself
Same idea!
Not working
I think it's not so good idea, better use format in print, like: print('%s%s' % (' ' * (n - i), '#' * i))
one liner
i have got the error as follows...
(lambda num: [print(('#'*x).rjust(num)) for x in range(num+1)])(int(input().strip())) ^ SyntaxError: invalid syntax
can you please help me why it so.....
Is there something wrong with this answer? Cannot pass the test case.
Bro...how the n value in rjust() is decreasing...first it will print n spaces than in next iteration what will happen..please explain
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.
Will Someone please explain me this python code ? I know how to do it using 2 loops
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
Thank you. I understood the code.
Your english is fine
try this
what is the main function of rjust?
right justification
n = int(input()) for i in range(n): print(' '(n-i-1)+'#'(i+1))
Python3 is even more love:
You can also do it pretty easy as a one-liner :)
Thanks :)
pls explain this code,cant understand.
I did it a little less concisely:
thanks)
Neat and great solution !!
Really it is very fantastic I was trying for i in range(n): k = n while k>i+1: print ' ', k -= 1 for j in range(i+1): print '#', print
Thank you
similar to my Scala code. I guess Scala creators borrowed from Java, Python, and Haskell quite a few ideas.
you don't need the second argument in rjust, default is space already ;)
my output is same but still showing fail, can anybody tell me whats wrong:
show me your code?
a=int(input())+1 i=0 while(a!=0): print(" "*a,end="") print("#"*i) a=a-1 i=i+1
didi u find anything
Could you please format your code or do one thing if its not in java then cross verify your solution with mine one. It will help you to figure out the mistake at your own.
Here is the video explanation –
https://youtu.be/gDltV7pJw7A
I'm beginner in python. Can you explain me this code
simpler one count = 1 while(count <= n): print(('#' * count).rjust(n)) count += 1
Good Evening! I was amazed at your succinct and easy solution to this problem. I am currently getting my feet wet in Python and have been using Hackerrank in order to improve my familiarity with it along with its online documentation.
I main question is this: How did you know that you could use print('#'*1) I have scoured programiz.com and Stack Overflow for ideas on handling this problem, but this technique never came up.
Any lead on becoming a better python programmer would be greatly appreciated!
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!
thank you
why we used rjust()?
This is how I did it.
good one!
Nice... Short and sweet..
IMHO the best solution
good
good one bro
Bravo bravo!
great stuff
great!
s it.
superu lol
This very good!
why
Thanks BlueBird this really helped me a lot
Why so much tables? One is enough:)
Ah, now this is what I call simple! Builder just confused me more. Thanks!
This is the best answer. Done in N time!
I really enjoyed this solution!
how the builder.length() - j works at first time?
nice one , simpler one
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).
I dont't think that this is O(n). The .replace() scans the char[] and you call it twice inside the for loop. So I think it's O(n²).
nice method
O(n) algorithm here. Two nested for loops are O(n^2) algorithm.
I did it like this ...
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):
here is my solution:
I did exactly the same and I wonder if this is the best way...
plzz explain this