Josephus problem Solution with code and explanation

Josephus problem Solution-

Read Josephus Story -here

Problem Statement

Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle in a fixed direction.​
The task is to choose the safe place in the circle so that when you perform these operations starting from 1st place in the circle, you are the last one remaining and survive.

Example 1:

Input:
n = 3 k = 2
Output: 3
Explanation: There are 3 persons so 
skipping 1 person i.e 1st person 2nd 
person will be killed. Thus the safe 
position is 3

Josephus problem Solution -

Josephus servived  because he knew maths and recursion that's why he stand at 40th place.

I also know recursion so I can also survive when I will face a similar problem So you should also know How to survive in such situation for that you also have to learn recursion.

😆😆😆XD

 i have to think how can i break 
this problem into subproblem 
People are standing in a circle so counting form very first place 
i have to delete every k-1 th node from current index
For example, if n = 5 and k = 2, then the safe position is 3. Firstly, 
the person at position 2 is killed, then person at position 4 is killed,
 then person at position 1 is killed. Finally, the person at position 5
 is killed. So the person at position 3 survives.
If n = 7 and k = 3, then the safe position is 4. The persons at positions 3, 6, 2, 7, 5, 1 are
 killed in order, and person at position 4 survives.

i have simple approach
i will first add all person in a vector and delete i will kill every 
k-1 th person and start countin again from new position

Now think about the smallest problem 
when I have only one person in my vector this will my solution no matter what is the value of k 

so let's write base case 

if(v.size()==0){

return v[0];

}

Now Think about induction step

If i am standing in a generic position in my vactor what will my intension 

to delete kth node from my posion if it is out of box start count from begin .

so i will find next index which i have to delete and then i will call my solve funcion which will return 

me final ans

induction step 

index=(index+k)%v.size();
	v.erase(v.begin()+index);
	return solve(v,index,k,ans);

final code is here- must share with your friends