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Question

int f(int n)

{

static int r=0;

if(n<=0)

return 1 ;

if (n>3)

{

r=n;

return f(n-2) +2;

}

return f(n-1)+r;

}

What is the value of f(5)?

 

Kindly explain how the recursion is working here along with the solution.

Comments

18

---

.....

Answer 1

f(5)

n=5 r=0

---

n=5,r=5

n>3

so,

f(3)+2

---

n=3

so,

f(2)+r

f(2)+5

---

f(1)+5

---

f(0)+5

---

f(0)=1

so f(5)=18

Comments

since f(0)=1

so f(1)=f(0)+5=6

f(2)=f(1)+5=11

f(3)=f(2)+5=16

f(5)=f(3)+2=16+2=18

Answer 2

f(3)+2

f(2)+r

f(1)+r

f(0)+r

here one thing is to note that r is static variable when we call f(5) r=n =>r=5

now 5>3 so f(5)=f(3)+2=16+2=18

now f(3) will be called so f(3)=f(2)+r=11+5=16

now f(2) will be called so f(2)=f(1)+r=6+5=11

now f(1) will be called so f(1)=f(0)+r =>1+5=6

f(0)=1

so finally the value of f(5) is 18