## 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.Ã¢€‹

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.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 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 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

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.`

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

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.

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

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

index=(index+k)%v.size(); | |

v.erase(v.begin()+index); | |

return solve(v,index,k,ans); |

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