## Josephus problem Solution-

### 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.Ã¢€‹
The task is to choose the safe place in the circle so that when you perform these operations starting from 1
st 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.ðŸ˜†ðŸ˜†ðŸ˜†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 indexFor 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 stepIf 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 ansinduction 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

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