Practice Recursion Medium (original) (raw)
Last Updated : 11 Apr, 2026
**Question 1 : Predict the output of the following program. What does the following fun2() do in general?
C++ `
using namespace std; #define LIMIT 1000 void fun2(int n) { if (n <= 0) return; if (n > LIMIT) return; cout << n << " "; fun2(2 * n); cout << n << " "; }
// Driver code int main() { fun2(100); }
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); }
// Driver code int main() { fun2(100); }
Java
class GFG { static int LIMIT = 1000; static void fun2(int n) { if (n <= 0) return; if (n > LIMIT) return;
System.out.print(n + " ");
fun2(2 * n);
System.out.print(n + " ");
}
// Driver code
public static void main(String[] args) { fun2(100); }}
Python
LIMIT = 1000
def fun2(n): if (n <= 0): return if (n > LIMIT): return print(n, end=" ") fun2(2 * n) print(n, end=" ")
Driver code
fun2(100)
C#
using System; class GFG {
static int LIMIT = 1000;
static void fun2(int n)
{
if (n <= 0)
return;
if (n > LIMIT)
return;
Console.Write(n + " ");
fun2(2 * n);
Console.Write(n + " ");
}
// Driver code
static void Main(string[] args) { fun2(100); }}
JavaScript
let LIMIT = 1000; function fun2(n) { if (n <= 0) return; if (n > LIMIT) return;
console.log(n + " ");
fun2(2 * n);
console.log(n + " ");}
// Driver code fun2(100)
`
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.
**Question 2 : Predict the output of the following program. What does the following fun() do in general?
C++ `
using namespace std;
int fun(int a[], int n) { int x; if (n == 1) return a[0]; else x = fun(a, n - 1); if (x > a[n - 1]) return x; else return a[n - 1]; }
// Driver code int main() { int arr[] = {12, 10, 30, 50, 100}; cout << " " << fun(arr, 5) << " "; getchar(); return 0; }
C
#include<stdio.h> int fun(int a[],int n) { int x; if(n == 1) return a[0]; else x = fun(a, n-1); if(x > a[n-1]) return x; else return a[n-1]; }
// Driver code int main() { int arr[] = {12, 10, 30, 50, 100}; printf(" %d ", fun(arr, 5)); getchar(); return 0; }
Java
class GFG {
static int fun(int a[],int n)
{
int x;
if(n == 1)
return a[0];
else
x = fun(a, n - 1);
if(x > a[n - 1])
return x;
else
return a[n - 1];
}
// Driver code
public static void main (String[] args)
{
int arr[] = {12, 10, 30, 50, 100};
System.out.println(" "+fun(arr, 5)+" ");
}}
Python
def fun( a, n): if(n == 1): return a[0] else: x = fun(a, n - 1) if(x > a[n - 1]): return x else: return a[n - 1]
Driver code
arr = [12, 10, 30, 50, 100] print(fun(arr, 5))
C#
using System;
public class GFG{ static int fun(int[] a,int n) { int x; if(n == 1) return a[0]; else x = fun(a, n - 1); if(x > a[n - 1]) return x; else return a[n - 1]; }
// Driver code
static public void Main ()
{
int[] arr = {12, 10, 30, 50, 100};
Console.Write(" "+fun(arr, 5)+" ");
}}
JavaScript
function fun(a, n) { var x; if(n == 1) return a[0]; else x = fun(a, n - 1); if(x > a[n - 1]) return x; else return a[n - 1]; }
// Driver code var arr = [12, 10, 30, 50, 100]; console.log(fun(arr, 5));
`
Answer: fun() returns the maximum value in the input array a[] of size n.
**Question 3 : Predict the output of the following program. What does the following fun() do in general?
C++ `
using namespace std;
int fun(int i) { if (i % 2) return (i++); else return fun(fun(i - 1)); }
// Driver code int main() { cout << " " << fun(200) << " "; getchar(); return 0; }
C
int fun(int i) { if ( i%2 ) return (i++); else return fun(fun( i - 1 )); }
// Driver code int main() { printf(" %d ", fun(200)); getchar(); return 0; }
Java
class GFG { static int fun(int i) { if (i % 2 == 1) return (i++); else return fun(fun(i - 1)); }
// Driver code
public static void main (String[] args) {
System.out.println(" " + fun(200) + " ");
}}
Python
def fun(i) :
if (i % 2 == 1) :
i += 1
return (i - 1)
else :
return fun(fun(i - 1))Driver code
print(fun(200))
C#
using System;
class GFG {
static int fun(int i)
{
if (i % 2 == 1)
return (i++);
else
return fun(fun(i - 1));
}
// Driver code
static public void Main()
{
Console.WriteLine(fun(200));
}}
JavaScript
function fun(i) { if (i % 2 == 1) return (i++); else return fun(fun(i - 1)); }
// Driver code console.log(fun(200));
`
Answer: If n is odd, then return n, else returns (n-1). Eg., for n = 12, you get 11 and for n = 11 you get 11. The statement _"return i++;" returns the value of i only as it is a post-increment.
**Question 4 : Predict the output of the following program. What does the following fun() do in general?
C++ `
using namespace std;
int fun(int n, int* fp) { int t, f;
if (n <= 2) {
*fp = 1;
return 1;
}
t = fun(n - 1, fp);
f = t + *fp;
*fp = t;
return f;}
// Driver code int main() { int x = 15; cout << fun(5, &x) << endl; return 0; }
C
int fun(int n, int* fp) { int t, f;
if (n <= 2) {
*fp = 1;
return 1;
}
t = fun(n - 1, fp);
f = t + *fp;
*fp = t;
return f;}
// Driver code int main() { int x = 15; printf("%d\n", fun(5, &x));
return 0;}
Java
class GFG { static int fp = 15; static int fun(int n) { int t, f;
if (n <= 2) {
fp = 1;
return 1;
}
t = fun(n - 1);
f = t + fp;
fp = t;
return f;
}
public static void main(String[] args)
{
System.out.println(fun(5));
}}
Python
fp = 15
def fun(n): global fp if (n <= 2): fp = 1 return 1
t = fun(n - 1)
f = t + fp
fp = t
return fDriver code
print(fun(5))
C#
using System;
class GFG { static int fp = 15; static int fun(int n) { int t, f;
if (n <= 2) {
fp = 1;
return 1;
}
t = fun(n - 1);
f = t + fp;
fp = t;
return f;
}
// Driver code
static public void Main() { Console.Write(fun(5)); }}
JavaScript
var fp = 15; function fun( n ) { var t, f;
if ( n <= 2 )
{
fp = 1;
return 1;
}
t = fun ( n - 1 );
f = t + fp;
fp = t;
return f;}
// Driver code console.log(fun(5))
`
The program calculates n-th Fibonacci Number. The statement t = fun ( n-1, fp ) gives the (n-1)th Fibonacci number and *fp is used to store the (n-2)th Fibonacci Number. The initial value of *fp (which is 15 in the above program) doesn't matter. The following recursion tree shows all steps from 1 to 10, for the execution of fun(5, &x).
(1) fun(5, fp)
/ \
(2) fun(4, fp) (8) t = 3, f = 5, *fp = 3
/ \
(3) fun(3, fp) (7) t = 2, f = 3, *fp = 2
/ \
(4) fun(2, fp) (6) t = 1, f = 2, *fp = 1
/
(5) *fp = 1 **Question 5 : Predict the output of the following program.
C++ `
using namespace std;
int minIndex(int arr[], int s, int e) { int sml = INT32_MAX; int mindex; for (int i = s; i < e; i++) { if (sml > arr[i]) { sml = arr[i]; mindex = i; } } return mindex; }
void fun2(int arr[], int start_index, int end_index) { if (start_index >= end_index) return; int min_index; int temp;
// minIndex() returns index of minimum value in
// array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index);
// swap the element at start_index and min_index
temp = arr[start_index];
arr[start_index] = arr[min_index];
arr[min_index] = temp;
fun2(arr, start_index + 1, end_index);}
// Driver code
int main() { int arr[] = {64, 25, 12, 22, 11}; int n = sizeof(arr) / sizeof(arr[0]);
fun2(arr, 0, n);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
return 0;}
C
// minimum index finder #include <stdio.h> #include <limits.h> int minIndex(int arr[], int s, int e) { int sml = INT_MAX; int mindex; for (int i = s; i < e; i++) { if (sml > arr[i]) { sml = arr[i]; mindex = i; } } return mindex; }
void fun2(int arr[], int start_index, int end_index) { if (start_index >= end_index) return; int min_index; int temp;
// minIndex() returns index of minimum value in
// array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index);
temp = arr[start_index];
arr[start_index] = arr[min_index];
arr[min_index] = temp;
fun2(arr, start_index + 1, end_index);}
// Driver code
int main() { int arr[] = {64, 25, 12, 22, 11}; int n = sizeof(arr) / sizeof(arr[0]); fun2(arr, 0, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); printf("\n");
return 0;}
Java
static int minIndex(int arr[], int s, int e) { int sml = Integer.MAX_VALUE; int mindex = s; for (int i = s; i < e; i++) { if (sml > arr[i]) { sml = arr[i]; mindex = i; } } return mindex; }
static void fun2(int arr[], int start_index, int end_index) { if (start_index >= end_index) return; int min_index; int temp;
// minIndex() returns index of minimum value in
// array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index);
temp = arr[start_index];
arr[start_index] = arr[min_index];
arr[min_index] = temp;
fun2(arr, start_index + 1, end_index);}
// Driver code
public static void main(String[] args) { int arr[] = {64, 25, 12, 22, 11}; int n = arr.length; fun2(arr, 0, n); for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); System.out.println(); }
Python
Minimum index finder
def minIndex(arr, s, e):
sml = sys.maxsize
mindex = 0
for i in range(s, e):
if (sml > arr[i]):
sml = arr[i]
mindex = i
return mindexdef fun2(arr, start_index, end_index):
if (start_index >= end_index):
return
# minIndex() returns index of minimum value in
# array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index)
arr[start_index], arr[min_index] = arr[min_index], arr[start_index]
fun2(arr, start_index + 1, end_index)Driver code
arr = [64, 25, 12, 22, 11] n = len(arr) fun2(arr, 0, n) print(*arr)
C#
using System;
class Recursion{
static int minIndex(int[] arr, int s, int e)
{
int sml = Int32.MaxValue;
int mindex = s;
for(int i = s; i < e; i++)
{
if(sml > arr[i])
{
sml = arr[i];
mindex = i;
}
}
return mindex;
}
static void fun2(int[] arr, int start_index, int end_index)
{
if(start_index >= end_index)
{
return;
}
int min_index;
int temp;
// minIndex() returns index of minimum value in
// array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index);
temp = arr[start_index];
arr[start_index] = arr[min_index];
arr[min_index] = temp;
fun2(arr, start_index + 1, end_index);}
// Driver code
static void Main(string[] args) { int[] arr = {64, 25, 12, 22, 11}; int n = arr.Length; fun2(arr, 0, n); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); Console.WriteLine(); } }
JavaScript
function minIndex(arr, s, e) { var sml = Number.MAX_SAFE_INTEGER; var mindex; for (let i = s; i < e; i++) { if (sml > arr[i]) { sml = arr[i]; mindex = i; } } return mindex; }
function fun2(arr, start_index, end_index) { if (start_index >= end_index) return; var min_index; var temp;
// minIndex() returns index of minimum value in
// array arr[start_index...end_index]
min_index = minIndex(arr, start_index, end_index);
// swap the element at start_index and min_index
temp = arr[start_index];
arr[start_index] = arr[min_index];
arr[min_index] = temp;
fun2(arr, start_index + 1, end_index);}
// Driver code
let arr = [64, 25, 12, 22, 11]; let n = arr.length; fun2(arr, 0, n); console.log(arr.join(" "));
`
**Answer: The function fun2() is a recursive implementation of Selection Sort.
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