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Josephus problem | Set 1 (A O(n) Solution)

  • Difficulty Level : Hard
  • Last Updated : 08 Nov, 2021

In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem. Following is the problem statement:

There are n people standing in a circle waiting to be executed. The counting out begins at some point in the circle and proceeds around the circle in a fixed direction. In each step, a certain number of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. 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. The task is to choose the place in the initial circle so that you are the last one remaining and so survive.
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.

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The problem has following recursive structure.  



  josephus(n, k) = (josephus(n - 1, k) + k-1) % n + 1
  josephus(1, k) = 1

After the first person (kth from beginning) is killed, n-1 persons are left. So we call josephus(n – 1, k) to get the position with n-1 persons. But the position returned by josephus(n – 1, k) will consider the position starting from k%n + 1. So, we must make adjustments to the position returned by josephus(n – 1, k). 

Following is simple recursive implementation of the Josephus problem. The implementation simply follows the recursive structure mentioned above.  

C++




#include <iostream>
using namespace std;
 
int josephus(int n, int k)
{
    if (n == 1)
        return 1;
    else
        /* The position returned by josephus(n - 1, k)
        is adjusted because the recursive call
        josephus(n - 1, k) considers the
        original position k % n + 1 as position 1 */
        return (josephus(n - 1, k) + k - 1) % n + 1;
}
 
// Driver Program to test above function
int main()
{
    int n = 14;
    int k = 2;
    cout << "The chosen place is " << josephus(n, k);
    return 0;
}
 
// This code is contributed by shubhamsingh10

C




#include <stdio.h>
 
int josephus(int n, int k)
{
    if (n == 1)
        return 1;
    else
        /* The position returned by josephus(n - 1, k) is
           adjusted because the recursive call josephus(n -
           1, k) considers the original position
           k%n + 1 as position 1 */
        return (josephus(n - 1, k) + k - 1) % n + 1;
}
 
// Driver Program to test above function
int main()
{
    int n = 14;
    int k = 2;
    printf("The chosen place is %d", josephus(n, k));
    return 0;
}

Java




// Java code for Josephus Problem
import java.io.*;
 
class GFG {
 
    static int josephus(int n, int k)
    {
        if (n == 1)
            return 1;
        else
            /* The position returned by josephus(n - 1, k)
            is adjusted because the recursive call
            josephus(n - 1, k) considers the original
            position k%n + 1 as position 1 */
            return (josephus(n - 1, k) + k - 1) % n + 1;
    }
 
    // Driver Program to test above function
    public static void main(String[] args)
    {
        int n = 14;
        int k = 2;
        System.out.println("The chosen place is "
                           + josephus(n, k));
    }
}
 
// This code is contributed by Prerna Saini

Python3




# Python code for Josephus Problem
 
 
def josephus(n, k):
 
    if (n == 1):
        return 1
    else:
 
        # The position returned by
        # josephus(n - 1, k) is adjusted
        # because the recursive call
        # josephus(n - 1, k) considers
        # the original position
        # k%n + 1 as position 1
        return (josephus(n - 1, k) + k-1) % n + 1
 
# Driver Program to test above function
 
 
n = 14
k = 2
 
print("The chosen place is ", josephus(n, k))
 
# This code is contributed by
# Sumit Sadhakar

C#




// C# code for Josephus Problem
using System;
 
class GFG {
 
    static int josephus(int n, int k)
    {
        if (n == 1)
            return 1;
        else
            /* The position returned
            by josephus(n - 1, k) is
            adjusted because the
            recursive call josephus(n
            - 1, k) considers the
            original position k%n + 1
            as position 1 */
            return (josephus(n - 1, k) + k - 1) % n + 1;
    }
 
    // Driver Program to test above
    // function
    public static void Main()
    {
        int n = 14;
        int k = 2;
        Console.WriteLine("The chosen "
                          + "place is " + josephus(n, k));
    }
}
 
// This code is contributed by anuj_67.

PHP




<?php
// PHP code for
// Josephus Problem
 
function josephus($n, $k)
{
    if ($n == 1)
        return 1;
    else
        /* The position returned by
           josephus(n - 1, k) is
           adjusted because the
           recursive call josephus
           (n - 1, k) considers the
           original position k%n + 1
           as position 1 */
        return (josephus($n - 1, $k) +
                    $k - 1) % $n + 1;
}
 
    // Driver Code
    $n = 14;
    $k = 2;
    echo "The chosen place is ", josephus($n, $k);
 
// This code is contributed by ajit.
?>

Javascript




<script>
 
    // Javascript code for Josephus Problem
     
    function josephus(n, k)
    {
        if (n == 1)
            return 1;
        else
            /* The position returned
            by josephus(n - 1, k) is
            adjusted because the
            recursive call josephus(n
            - 1, k) considers the
            original position k%n + 1
            as position 1 */
            return (josephus(n - 1, k)
                       + k-1) % n + 1;
    }
       
    let n = 14;
    let k = 2;
    document.write("The chosen " + "place is " + josephus(n, k));
     
</script>
Output
The chosen place is 13

Time Complexity: O(n)

Auxiliary Space: O(n)

Another Approach using List:

The simple approach is to create a list and add all values from 1 to n in it. Create a recursive function that takes list, start (position at which counting will start) and k ( number of person to be skipped) as argument. If size of list is one i.e. only one person left then return this position. Otherwise, start counting k person in clockwise direction from starting position and remove the person at kth position. Now the person at kth position is removed and now counting will start from this position. This process continues till only one person left.

pseudo-code :

Josephus( list , start , k){
   if list.size = 1
       return list[0]
   start = (start + k) % list.size
   list.remove( start )
   return Josephus( list, start, k)
}

C++ Code

C++




#include <bits/stdc++.h>
 
using namespace std;
 
void Josh(vector<int> person, int k, int index)
{
    // Base case , when only one person is left
    if (person.size() == 1) {
        cout << person[0] << endl;
        return;
    }
 
    // find the index of first person which will die
    index = ((index + k) % person.size());
 
    // remove the first person which is going to be killed
    person.erase(person.begin() + index);
 
    // recursive call for n-1 persons
    Josh(person, k, index);
}
 
int main()
{
    int n = 14; // specific n and k  values for original
                // josephus problem
    int k = 2;
    k--; // (k-1)th person will be killed
    int index
        = 0; // The index where the person which will die
 
    vector<int> person;
    // fill the person vector
    for (int i = 1; i <= n; i++) {
        person.push_back(i);
    }
 
    Josh(person, k, index);
}

C#




using System;
using System.Collections.Generic;
class GFG {
     
    static void Josh(List<int> person, int k, int index)
    {
        // Base case , when only one person is left
        if (person.Count == 1) {
            Console.WriteLine(person[0]);
            return;
        }
      
        // find the index of first person which will die
        index = ((index + k) % person.Count);
      
        // remove the first person which is going to be killed
        person.RemoveAt(index);
      
        // recursive call for n-1 persons
        Josh(person, k, index);
    }
 
  // Driver code
  static void Main()
  {
    int n = 14; // specific n and k  values for original
                // josephus problem
    int k = 2;
    k--; // (k-1)th person will be killed
    int index
        = 0; // The index where the person which will die
  
    List<int> person = new List<int>();
    // fill the person vector
    for (int i = 1; i <= n; i++) {
        person.Add(i);
    }
  
    Josh(person, k, index);
  }
}
 
// This code is contributd by divyesh072019.
Output



13

Time Complexity: O(n)

Example

Input : n = 5,  k = 2
output : 3

Explanation :

Add all values from 1 to n in the list. We will call the recursive function with start = 0 and k = 1 (0-indexing)

Now the element at 1-index (person number 2) will be killed. And it is removed from list. The new counting will begins from 1-index, person at 1-index killed so now person at 2-index (person number 3) comes to 1-index and counting starts from here now.

Now we have 4 person, counting start from 1-index (person number 3) and the person at kth (2-index ) position will be killed. 

The person at 2-index (person number 4) killed so now we have 3 person left and the person (person number 5) at 3-index shifted to 2-index. And counting starts from here.



The person at 0-index killed and we have now two person left in the circle. And the person at 1-index shifted to 0-index i.e. person number 3.

Final counting done and person at 1-index killed and only person who is left is at position 3.

This is a solution to the Josephus problem in C++.

For n = 47 and k = 5.

C++




#include <bits/stdc++.h>
using namespace std;
 
int Josephus(int, int);
 
int main()
{
    int n, k;
    cin >> n >> k;
    cout << Josephus(n, k);
    return 0;
}
 
int Josephus(int n, int k)
{
    k--;
    int arr[n];
    for (int i = 0; i < n; i++) {
        arr[i] = 1; // Makes all the 'n' people alive by
                    // assigning them value = 1
    }
    int cnt = 0, cut = 0,
        num = 1; // Cut = 0 gives the sword to 1st person.
    while (
        cnt
        < (n - 1)) // Loop continues till n-1 person dies.
    {
        while (num <= k) // Checks next (kth) alive persons.
        {
            cut++;
            cut = cut % n; // Checks and resolves overflow
                           // of Index.
            if (arr[cut] == 1) {
                num++; // Updates the number of persons
                       // alive.
            }
        }
        num = 1; // refreshes value to 1 for next use.
        arr[cut]
            = 0; // Kills the person at position of 'cut'
        cnt++; // Updates the no. of killed persons.
        cut++;
        cut = cut
              % n; // Checks and resolves overflow of Index.
        while (arr[cut]
               == 0) // Checks the next alive person the
                     // sword is to be given.
        {
            cut++;
            cut = cut % n; // Checks and resolves overflow
                           // of Index.
        }
    }
    return cut + 1; // Output is the position of the last
                    // man alive(Index + 1);
}
 
/********************THIS CODE IS PRESENTED BY SHISHANK
 * RAWAT**************************/

If you guys liked it, please comment and help me with further improvements.

Output
The safe position : 4

Josephus problem | Set 2 (A Simple Solution when k = 2)

Source: 
http://en.wikipedia.org/wiki/Josephus_problem

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 




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