LESSON 175 OF 348 · STRINGS

Print Triangle of Stars in Kotlin: A Loop That Grows a String

Practice O(n²) Read the full explanation

Given a positive integer n print a triangle of stars as in the example below

Example:

Input: n = 5
Output:
*
**
***
****
*****
Explanation: Print 5 lines, on ith line print i stars.

STEP 1 OF 4

Guided

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Print Triangle Of Stars in Kotlin, explained

Problem overview

The function printStars receives a positive integer n and prints a triangle of stars. It has no return value: its job is the output. Line number i (counting from 1) holds i stars, so the triangle grows by one star per line.

For n = 5 the output is:

*
**
***
****
*****

Approach: each row is the previous row plus one star

There are two ways to think about a row. You can build it from scratch each time, with an inner loop that prints i stars. Or you can notice that row 3 is just row 2 with a star added at the end. The second view needs no inner loop: keep one string, add a star, print it, and repeat.

Following n = 3: the string starts empty. After the first pass it is "*" and gets printed. After the second pass it is "**", and after the third "***". Each pass prints one line, so n passes print n lines.

Step by step

Declare str as an empty var. It holds the current row. It must be a var because the loop gives it a new value on every pass.
Loop once per row. for (i in 0 until n) runs the body exactly n times, since until excludes its upper bound. The counter i only counts the passes; the body never reads it.
Add a star. str += "*" makes a new string, the old one plus a star, and stores it back in str. Strings in Kotlin can't be edited, so += replaces the whole value.
Print the row. println(str) writes the string and ends the line. It sits inside the loop, after the star was added, so each pass prints the row it just extended.

The solution in Kotlin

class Solution {
    fun printStars(n: Int) {
        var str = ""
        for (i in 0 until n) {
            str += "*"
            println(str)
        }
    }
}

Kotlin details worth knowing:

println versus print. println adds a line break after the text, and print doesn't. Using print here would put every star on one long line.
0 until n versus 1..n. 0 until n gives 0, 1, ..., n - 1, and 1..n gives 1, 2, ..., n. Both run n times. The first is the usual choice when the counter isn't used.
An unused loop variable. Kotlin accepts it. An underscore is not allowed as a loop variable name in a for loop (the compiler reports that the name is reserved), so the variable keeps a real name. repeat(n) { ... } is an alternative that needs no counter.
The return type. A function without a declared return type returns Unit, Kotlin's counterpart of void. Nothing is returned, only printed.

Complexity

Let n be the input.

Time: O(n^2). Row i has i stars, and both appending and printing cost time proportional to the row length. Altogether the program handles 1 + 2 + ... + n = n(n + 1) / 2 stars. That is the size of the output itself, so no solution can do better than O(n^2).

Space: O(n). The only extra data is the current row, which has at most n stars.

Edge cases and common mistakes

n = 1. One pass, one line: *.
Printing outside the loop. After the loop, str holds the longest row only, so the output would be a single line of n stars.
Declaring str inside the loop. Each pass would start from an empty string, and every line would be a single star.
Adding the star after printing. The first line would be empty, and the last row would never print, so the triangle would be shifted by one.
Using val for str. str += "*" then fails to compile, because a val cannot be reassigned.
Off by one. 0 until n - 1 prints one row too few, and 0..n prints one too many.

Testing

n = 1 prints a single *.
n = 3 prints *, **, *** on three lines.
n = 5 prints the triangle shown above, five lines in total.

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