Can the print statement be placed after the recursive call?

Yes. In that case, all recursive calls are made first, and the names are printed while the calls return. Since the same string is printed every time, the visible output remains the same.

What happens when n = 0?

The base case is reached immediately, so the function prints nothing.

Why is name passed into every recursive call?

Each call needs access to the same string that must be printed. Passing name keeps the function independent of global variables.

Can this problem be solved using a loop?

Yes. A loop would use O(1) auxiliary space and would be more practical for this task. Recursion is used here to understand base cases, recursive calls, and call-stack behaviour.

Print Name N Times Using Recursion

Q: Why is n <= 0 used as the base case?

Once n reaches 0, the name has already been printed the required number of times. Using n <= 0 also stops the function safely when a negative value is passed.

Given a non-negative integer n and a string name, print the given string exactly n times using recursion.

Standard loops such as for and while must not be used.

Example 1

Input: n = 3, name = "Code"

Output: Code Code Code

Explanation: The function executes three separate times. During each execution, it outputs the given string exactly once, resulting in three consecutive prints.

Example 2

Input: n = 1, name = "Hello"

Output: Hello

Explanation: Because the target count is exactly 1, the string is printed once before the recursive sequence immediately terminates.

Practice it by yourself before moving onto the solutionTry now

Approach

A loop repeats the same task until a condition is met. Recursion can create the same repetition by letting the function call itself with a smaller value.

Here, every call prints the given name once. After printing, the function calls itself with n - 1, meaning one fewer print remains. Once n reaches 0, the required number of prints has been completed, so the recursion stops.

Algorithm

Define a recursive function that receives n and name. The value of n represents how many times the name is still left to be printed.
When n <= 0, return from the function because no more prints are required. This condition also prevents further calls for a negative value of n.
Print name once for the current recursive call.
Call the function again with n - 1 and the same name. Reducing n after every call brings the recursion closer to the base case.

Dry Run

Print Name N Times Dry Run .png

Solution

1#include <bits/stdc++.h>
2using namespace std;
3
4class Solution {
5public:
6    void printName(int n, const string& name) {
7        // Base case: no more prints are left.
8        if (n <= 0) {
9            return;
10        }
11
12        cout << name << '\n';
13
14        // One print is complete, so continue with n - 1.
15        printName(n - 1, name);
16    }
17};
18
19int main() {
20    int n = 3;
21    string name = "Striver";
22
23    Solution solution;
24    solution.printName(n, name);
25
26    return 0;
27}

Complexity Analysis

Time Complexity: O(N), because the function prints the name once for each value from N down to 1.

Space Complexity: O(N), because each recursive call occupies one call-stack frame until the base case is reached.

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