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# Practice Questions for Recursion | Set 3

Explain the functionality of below recursive functions.

Question 1

## C++

 `void` `fun1(``int` `n)  ` `{  ` `   ``int` `i = 0;    ` `   ``if` `(n > 1)  ` `     ``fun1(n - 1);  ` `   ``for` `(i = 0; i < n; i++)  ` `     ``cout << ``" * "``;  ` `}  ` ` `  `// This code is contributed by shubhamsingh10 `

## C

 `void` `fun1(``int` `n) ` `{ ` `   ``int` `i = 0;   ` `   ``if` `(n > 1) ` `     ``fun1(n-1); ` `   ``for` `(i = 0; i < n; i++) ` `     ``printf``(``" * "``); ` `} `

## Java

 `static` `void` `fun1(``int` `n)  ` `{  ` `   ``int` `i = ``0``;    ` `   ``if` `(n > ``1``)  ` `     ``fun1(n - ``1``);  ` `   ``for` `(i = ``0``; i < n; i++)  ` `     ``System.out.print(``" * "``);  ` `}  ` `  `  `// This code is contributed by shubhamsingh10 `

## Python3

 `def`  `fun1(n): ` `    ``i ``=` `0`  `    ``if` `(n > ``1``): ` `        ``fun1(n ``-` `1``)  ` `    ``for` `i ``in` `range``(n): ` `        ``print``(``" * "``,end``=``"") ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `static` `void` `fun1(``int` `n)  ` `{  ` `    ``int` `i = 0;  ` `    ``if` `(n > 1)  ` `        ``fun1(n-1);  ` `    ``for` `(i = 0; i < n; i++)  ` `        ``Console.Write(``" * "``); ` `}  ` ` `  `// This code is contributed by shubhamsingh10 `

## Javascript

 ` `

Answer: Total numbers of stars printed is equal to 1 + 2 + …. (n-2) + (n-1) + n, which is n(n+1)/2.

Time complexity: O(n)
Auxiliary Space:  O(n), due to recursion call stack

Question 2

## C++

 `#define LIMIT 1000 ` `void` `fun2(``int` `n) ` `{ ` `  ``if` `(n <= 0) ` `     ``return``; ` `  ``if` `(n > LIMIT) ` `    ``return``; ` `  ``cout << n <<``" "``; ` `  ``fun2(2*n); ` `  ``cout << n <<``" "``; ` `} ` ` `  `// This code is contributed by shubhamsingh10 `

## C

 `#define LIMIT 1000 ` `void` `fun2(``int` `n) ` `{ ` `  ``if` `(n <= 0) ` `     ``return``; ` `  ``if` `(n > LIMIT) ` `    ``return``; ` `  ``printf``(``"%d "``, n); ` `  ``fun2(2*n); ` `  ``printf``(``"%d "``, n); ` `}    `

## Java

 `int` `LIMIT = ``1000``; ` `void` `fun2(``int` `n) ` `{ ` `    ``if` `(n <= ``0``) ``return``; ` `    ``if` `(n > LIMIT) ``return``; `   `    ``System.out.print(String.format(``"%d "``, n)); ` `    ``fun2(``2` `* n); ` `    ``System.out.print(String.format(``"%d "``, n)); ` `} `

## Python3

 `LIMIT ``=` `1000` `def` `fun2(n): ` `    ``if` `(n <``=` `0``): ` `        ``return` `    ``if` `(n > LIMIT): ` `        ``return` `    ``print``(n, end``=``" "``) ` `    ``fun2(``2` `*` `n) ` `    ``print``(n, end``=``" "``) ` ` `  `# This code is contributed by shubhamsingh10 `

## C#

 `int` `LIMIT = 1000 ` `void` `fun2(``int` `n) ` `{ ` `    ``if` `(n <= 0) ` `        ``return``; ` `    ``if` `(n > LIMIT) ` `        ``return``; ` `    ``Console.Write(n+``" "``); ` `    ``fun2(2*n); ` `    ``Console.Write(n+``" "``); ` `} ` ` `  `// This code is contributed by Shubhamsingh10 `

## Javascript

 ` `

Answer: For a positive n, fun2(n) prints the values of n, 2n, 4n, 8n … while the value is smaller than LIMIT. After printing values in increasing order, it prints same numbers again in reverse order. For example fun2(100) prints 100, 200, 400, 800, 800, 400, 200, 100.
If n is negative, the function is returned immediately.

Time complexity: O(n)
Auxiliary Space:  O(log2n), due to recursion call stack