CODEWITHSUNDEEP
java
user1693251
user1693251

Reputation:

Is there any way to make this more efficient/reduce the amount of loops?

Given an int n, print a staircase using #. This is from hacker rank, staircase problem. example: n = 4.

output:

   #
  ##
 ###
####

whereas each row has the same amount of columns but the # signs increase and the space decrease as we keep going through the rows.

I've solved the problem, just trying to see if there is a more efficient way

public static void staircase(int n) {
    int spaceCounter = 0;

    for(int i = 1; i <= n; i++) { // Takes care of the rows
        spaceCounter = n - i;

        // Takes care of the column by printing a space until a # sign is required then it would print so.
        for (int j = 1; j <= spaceCounter; j++) {
            System.out.print(" ");

            if (j == spaceCounter) {
                //Prints as many #s as needed (n minus the number of spaces needed)
                for(int k = 1; k <= (n - spaceCounter); k++) {
                    System.out.print("#");
                }

               //makes sure it goes to the next life after being done with each row
                System.out.println();
            }
        }

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

Upvotes: 1

Views: 98

Answers (1)

Jacob G.
Jacob G.

Reputation: 29730

Using Java 11, you can utilize String#repeat for an efficient solution that uses a single for-loop:

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

All we do is calculate the amount of spaces that are needed for the specific line, and then the number of # characters needed is simply n minus the amount of spaces used.

If n is a large value, you can build a String (using a StringBuilder) and then print it instead of calling System.out.println n times:

public static void staircase(int n) {
    var sb = new StringBuilder();

    for (int i = 1; i <= n; i++) {
        sb.append(" ".repeat(n - i)).append("#".repeat(i)).append('\n');
    }

    System.out.print(sb);
}

Upvotes: 8

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