PPractick

The problem

A classroom display counts up from 1 to N on screen, and the exercise is to generate the sequence recursively, without a loop. It's the classic problem for learning that the position of the recursive call matters. Printing before the recursive call produces one order, and printing after it (on the way back up the call stack) produces the reverse. Here you'll write both versions, one that counts up by carrying an index forward and one that reaches 1 first and prints while unwinding, and understand why both print 1 to N.

Example: N = 5 → 1 2 3 4 5.

Function signature

void printForward(int i, int n);  // print i, then recurse on i + 1 (call as printForward(1, n))
void printByBacktracking(int n);  // recurse on n - 1 first, then print n

Test cases

Input n	Output	Explanation
5	1 2 3 4 5	Both versions
1	1
0	(nothing)	Base case immediately
10	1 2 3 4 5 6 7 8 9 10

(Confirm: n ≥ 0; numbers separated by spaces; no loops.)

How interviewers phrase it

• "Print numbers from 1 to N without using a loop."
• "Can you do it using only one parameter?" (This forces the backtracking version.)

Clarifying questions to ask

1. Can I use an extra parameter?
2. How large can N be? (Recursion depth.)
3. What separator between numbers?

Follow-ups to expect

1. "Why does printing after the recursive call still give 1 to N?"
2. "What's the difference between the two approaches?"
3. "What happens for N = 10^6?"
4. "Complexity?"

Related questions

• Understand recursion by print something N times Easy
• Print name N times using recursion Easy
• Print N to 1 using Recursion Easy
• Sum of First N Numbers Easy
• Factorial of a given number Easy
• Reverse an array Easy

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