How recursion stack works

#1

my i/p is n=4,k=3.

f(-2,3) <- this will return 0;
f(1,3) <- whats the ans of rest of the steps ?
f(3,3)
f(4,3)

#2

hello anyone herer ?

#3

first you put f(n) in the stack,
Bhaiya has taught this thing in an intuitive/graphic way.
let me explain this in a technical way.

push f(n) to the stack
repeat until !stack.empty()
f(i) = stack.top();
stack.pop();
if arrived at base case:
    add it to the ans
else:
    push f(i-1)...f(i-k) in the stack

f(4) = f(3) + f(2) + f(1)
f(3) = f(2) + f(1) + f(0)
f(2) = f(1) + f(0) + f(-1)
f(1) = 1
f(0) = 1

#4

what is wrong in this ?
f(-2,3) <- this will return 0;
f(1,3) <- whats the ans of rest of the steps ?
f(3,3)
f(4,3)

#5

can you explain with the stack i have made

#6

Where is f(2,3)?..

#7

n=4,k=3
f(-2,3) <- -----------> (4-3,3)
f(1,3) <- -----------> (4-2,3)
f(3,3) -----------> (4-1,3)
f(4,3)
there is no place for f(2,3) ? why should we include that ?

#8

I really can’t understand what you are trying to say here.
4-1 = 3 ok that’s fine
but then you do 4-2 = 1 ?

#9

f(-2,3)
f(1,3)
f(3,3)
f(4,3)

#10

NOW TELL ME WHAT RETURNS WHAT VALUE ?

#11

f(1,3) would add f(1-1,3) , f(1-2,3) , f(1-3,3) to the stack.
f(0,3) would return 1;
the other two would return zero.

#12

f(1,3) would add f(1-1,3) , f(1-2,3) , f(1-3,3) to the stack. how ?
i dont understand this line

#13

just to make sure we are on the same page, can you please tell me the recurrence relation of the problem.

#14

i am skipping this problem i dont want to do this anymore thanks

#15

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