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LESSON 132 OF 415 · LOOPS & CONDITIONALS

Print Even and Odd in Swift: for Loop with if and else

Practice Read the full explanation

Given a positive integer n, for each number from 0 to n - 1, print "odd" if the number is odd or the number itself if it is even.

Example:

Input: n = 6
Output (console):
0
odd
2
odd
4
odd

Note: You have to print to the console, your function shouldn't return anything.

STEP 1 OF 3
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    SWIFT · LOOPS & CONDITIONALS

    Print Even Odd in Swift, explained

    Problem overview

    The method printEvenOdd(_ n: Int) receives a positive integer n. For every number from 0 to n - 1 it prints one line: the word odd if the number is odd, or the number itself if it is even. The method returns nothing.

    For n = 6 the numbers are 0, 1, 2, 3, 4 and 5, and the output is:

    0
odd
2
odd
4
odd

    Approach: loop, then choose between two outputs

    Each number needs the same treatment: test whether it is odd, and print one of two things. That is a loop that visits the numbers 0 to n - 1, with an if and an else inside it. The test uses the remainder operator: a number divided by 2 leaves remainder 1 when it is odd and 0 when it is even.

    Step by step

    1. Loop with a half-open range. for i in 0..<n visits 0, 1, ..., n - 1. The end n is left out, which is exactly the range the task asks for.
    2. Test for odd. Inside the loop, an if checks i % 2 == 1. If it is true, print the string "odd".
    3. Handle the rest with else. A number that is not odd is even, so the else branch prints i. Exactly one of the two branches runs on every pass.

    A short trace

    For n = 3 the loop runs with i equal to 0, 1 and 2. At 0 the remainder is 0, so the test fails and the else branch prints 0. At 1 the remainder is 1, so the method prints odd. At 2 the remainder is 0 again and it prints 2. The range ends before 3, so the loop stops.

    The solution in Swift

    class Solution {
    func printEvenOdd(_ n: Int) {
        for i in 0..<n {
            if i % 2 == 1 {
                print("odd")
            } else {
                print(i)
            }
        }
    }
}

    Swift details worth knowing:

    • 0..<n versus 0...n. The closed range would also print a line for n itself, one line too many.
    • Two kinds of output. print("odd") prints a string and print(i) prints an Int. Both calls are fine, because print accepts any value, and both end the line.
    • Testing for 1. Because the loop only sees non-negative numbers, the remainder is 0 or 1. With negative numbers, -3 % 2 would be -1 in Swift, so a test for == 1 would miss negative odd numbers. Here that cannot happen.
    • The loop variable i is a constant created by the loop, and it is visible only inside it.
    • When n is 0. The range 0..<0 is empty and valid, so nothing is printed. A negative n would crash, as the range would have its end below its start, but the task guarantees a positive n.

    Complexity

    Time: O(n). The loop runs n times and each pass does constant work.

    Space: O(1). No data structure grows with n.

    Edge cases and common mistakes

    • n = 1. Only 0 is visited, so the output is a single line, 0.
    • Printing the quotes. print("odd") prints odd without the quotation marks, which is what is wanted.
    • Forgetting the else. Without it, even numbers would print nothing, since the only print is inside the if.
    • Using = for comparison. i % 2 = 1 is a compile error. Use ==.
    • Starting at 1. A loop from 1 would skip 0 and print a wrong first line.

    Testing

    • printEvenOdd(6) prints 0, odd, 2, odd, 4, odd.
    • printEvenOdd(1) prints 0.
    • printEvenOdd(3) prints 0, odd, 2.
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