LESSON 127 OF 338 · LOOPS & CONDITIONALS

Print Even Odd in Go: if and else Inside a For Loop

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

Guided

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Print Even Odd in Go, explained

Problem overview

The function printEvenOdd takes a positive integer n. For every number from 0 to n - 1, it prints one line: the word odd when the number is odd, and the number itself when it is even. Nothing is returned.

For n = 6 the numbers are 0 to 5. The output is 0, odd, 2, odd, 4, odd, one item per line. The loop never reaches 6, because the range stops at n - 1.

Approach: one loop, one decision per number

Every number in the range needs exactly one line of output, and the line depends on a yes-or-no question: is the number odd? That is a loop with an if/else inside it. Both branches print, so each pass produces exactly one line.

To tell odd from even, use the remainder after dividing by 2. Odd numbers leave remainder 1 and even numbers leave 0. For the non-negative numbers here, testing for remainder 1 is enough.

Step by step

Loop over the range. A for loop with a variable i that starts at 0 and continues while i < n covers exactly 0, 1, ..., n - 1. The post statement is i++.
Handle the odd numbers. Inside the loop, an if checks i % 2 == 1. In that branch, fmt.Println("odd") prints the word.
Handle the rest. The else branch runs for the even numbers, and fmt.Println(i) prints the number itself. Note the difference between the two prints: one has quotes around a word, the other has the variable.

The solution in Go

func printEvenOdd(n int) {
    for i := 0; i < n; i++ {
        if i % 2 == 1 {
            fmt.Println("odd")
        } else {
            fmt.Println(i)
        }
    }
}

The Go details:

else sits on the same line as the closing brace. Writing } and then else on the next line is a syntax error, because Go inserts a semicolon at the end of the line.
fmt.Println accepts different types. The same function prints a string and an int, so both branches can use it.
No ternary operator. Go has no condition ? a : b, so a two-way choice is written as if/else.
The remainder operator % works only on integers, and there is no automatic conversion in Go, so i and 2 are both int here.

Following the loop by hand

With n = 4, the loop makes four passes. For i = 0, the remainder is 0, so the else branch prints 0. For i = 1, the remainder is 1, so the word is printed. For i = 2, the number 2 is printed, and for i = 3, the word again. Then i becomes 4, 4 < 4 is false and the loop stops. Each pass takes exactly one of the two branches, never both and never neither, which is why the output has exactly n lines.

Complexity

Time: O(n). There is one pass per number, and each pass does a constant amount of work.

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

Edge cases and common mistakes

Zero is even. 0 % 2 is 0, so the first line of output is the number 0, not the word.
The upper bound. i <= n would add one extra line for n itself. With n = 6 that adds a final 6.
Quotes mixed up. Printing "i" in the else branch gives the letter instead of the number, and printing odd without quotes does not compile, because Go looks for a variable called odd.
Two separate if statements. Using one for odd and another for even also works, but else says the two cases are opposites and is shorter.
n = 1. The only number is 0, so the output is a single line, 0.

Testing

n = 6 prints 0, odd, 2, odd, 4, odd.
n = 1 prints 0.
n = 2 prints 0, then odd.
n = 4 prints 0, odd, 2, odd.

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