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Josephus Problem

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  • Difficulty Level : Medium
  • Last Updated : 07 Dec, 2022
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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 the kth person is killed in a circle. The task is to choose the person in the initial circle that survives.

Examples:

Input: N = 5 and k = 2
Output: 3
Explanation: Firstly, the person at position 2 is killed, 
then the person at position 4 is killed, then the person at position 1 is killed. 
Finally, the person at position 5 is killed. So the person at position 3 survives. 

Input: N = 7 and k = 3
Output: 4
Explanations: The persons at positions 3, 6, 2, 7, 5, and 1 are killed in order, 
and the person at position 4 survives.

Josephus problem using List

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

Pseudocode :

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

Follow the below steps to Implement the idea:

  • Create a vector person and push all the values from 1 to N in person.
  • Recursively eliminate the index element 
    • Erase the element on the position index.
    • Call for (index + k)% size of person. 
    • If size of person = 1, return person[i].

Below is the Implementation of the above approach:

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);
}

Java




import java.util.*;
 
class GFG{
 
 
  static void Josh(List<Integer> person, int k, int index)
  {
     
    // Base case , when only one person is left
    if (person.size() == 1) {
      System.out.println(person.get(0));
      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.remove(index);
 
    // recursive call for n-1 persons
    Josh(person, k, index);
  }
 
  // Driver code
  public static void main(String [] args)
  {
    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<Integer> person = new ArrayList<>();
     
    // fill the person vector
    for (int i = 1; i <= n; i++) {
      person.add(i);
    }
 
    Josh(person, k, index);
  }
}
 
// This code is contributed by umadevi9616

Python3




# Python code for Josephus Problem
def Josh(person, k, index):
   
  # Base case , when only one person is left
  if len(person) == 1:
    print(person[0])
    return
   
  # find the index of first person which will die
  index = ((index+k)%len(person))
   
   # remove the first person which is going to be killed
  person.pop(index)
   
  # recursive call for n-1 persons
  Josh(person,k,index)
 
# Driver Program to test above function
n = 14 # specific n and k  values for original josephus problem
k = 2
k-=1   # (k-1)th person will be killed
 
index = 0
 
# fill the person vector
person=[]
for i in range(1,n+1):
  person.append(i)
 
Josh(person,k,index)
 
# This code is contributed by
# Gaurav Kandel

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 contributed by divyesh072019.

Javascript




<script>
    function Josh( person , k , index) {
 
        // Base case , when only one person is left
        if (person.length == 1) {
            document.write(person[0]);
            return;
        }
 
        // find the index of first person which will die
        index = ((index + k) % person.length);
 
        // remove the first person which is going to be killed
         if (index > -1) {
       person.splice(index, 1);
    }
 
        // recursive call for n-1 persons
        Josh(person, k, index);
    }
 
    // Driver code
     
        var n = 14; // specific n and k values for original
        // josephus problem
        var k = 2;
        k--; // (k-1)th person will be killed
        var index = 0; // The index where the person which will die
 
        var person = [];
 
        // fill the person vector
        for (var i = 1; i <= n; i++) {
            person.push(i);
        }
 
        Josh(person, k, index);
         
// This code is contributed by umadevi9616
</script>

Output

13

Time Complexity: O(N2)
Auxiliary Space: O(N), For recursion stack

Approach to solve Josephus problem iteratively:

Illustration:

N = 5,  k = 2

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 the list. The new counting will begin from 1-index, the 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 people, counting starting 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) was killed so now we have 3 people left and the person (person number 5) at 3-index shifted to 2-index. And counting starts from here.

The person at the 0-index was 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 the person at 1-index killed and the only person who is left is at position 3.

Follow the below steps to Implement the idea:

  • Initialize variables num, cnt, and cut with 1, 0, and 0 respectively and an array arr[] of size N with the initial value set as 1.
  • Run a while loop till cnt < N:
    • Run a while loop till num is less than equal to k.
      • Increment cut by one and take modulo by N
      • If arr[cut] = 1 increment num by one.
    •  Set num = 1, arr[cut] = 0 and increment cnt and cut by one and cut = cut % n;
    • Run a while loop till arr[cut] = 0 and increment cut by one.
  • Return cnt + 1 as the required answer.

Below is the Implementation of the above approach:

C++




#include <bits/stdc++.h>
using namespace std;
 
int Josephus(int, int);
 
int Josephus(int n, int k)
{
    k--;
    int arr[n];
 
    // Makes all the 'n' people alive by
    // assigning them value = 1
    for (int i = 0; i < n; i++) {
        arr[i] = 1;
    }
    int cnt = 0, cut = 0,
        // Cut = 0 gives the sword to 1st person.
        num = 1;
 
    // Loop continues till n-1 person dies.
    while (cnt < (n - 1)) {
 
        // Checks next (kth) alive persons.
        while (num <= k) {
            cut++;
 
            // Checks and resolves overflow
            // of Index.
            cut = cut % n;
            if (arr[cut] == 1) {
                // Updates the number of persons
                // alive.
                num++;
            }
        }
 
        // Refreshes value to 1 for next use.
        num = 1;
 
        // Kills the person at position of 'cut'
        arr[cut] = 0;
 
        // Updates the no. of killed persons.
        cnt++;
        cut++;
 
        // Checks and resolves overflow of Index.
        cut = cut % n;
 
        // Checks the next alive person the
        // sword is to be given.
        while (arr[cut] == 0) {
            cut++;
 
            // Checks and resolves overflow
            // of Index.
            cut = cut % n;
        }
    }
 
    // Output is the position of the last
    // man alive(Index + 1);
    return cut + 1;
}
 
// Driver code
int main()
{
    int n = 14, k = 2;
    cout << Josephus(n, k);
    return 0;
}
 
// THIS CODE IS PRESENTED BY SHISHANK RAWAT

Java




// Java code to implement the above approach
import java.io.*;
class GFG {
 
    public static void main(String[] args)
    {
        int n = 14, k = 2;
        System.out.println(Josephus(n, k));
    }
 
    public static int Josephus(int n, int k)
    {
        k--;
        int arr[] = new int[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 contributed by Shubham Singh

Python3




def Josephus(n, k):
     
    k -= 1
    arr = [0]*n
    for i in range(n):
        arr[i] = 1 # Makes all the 'n' people alive by
        # assigning them value = 1
    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 += 1
            cut = cut % n # Checks and resolves overflow
            # of Index.
            if (arr[cut] == 1):
                num+=1 # 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 += 1 # Updates the no. of killed persons.
        cut += 1
        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 += 1
            cut = cut % n # Checks and resolves overflow
            # of Index.
     
    return cut + 1 # Output is the position of the last
                    # man alive(Index + 1)
 
# Driver Code
n, k = 14, 2 #map (int, input().splut())
print(Josephus(n, k))
 
# This code is contributed by ShubhamSingh

C#




// C# code to implement the above approach
using System;
using System.Linq;
 
public class GFG{
 
  public static void Main ()
  {
    int n = 14, k = 2;
    Console.Write(Josephus(n, k));
  }
 
  public static int Josephus(int n, int k)
  {
    k--;
    int[] arr = new int[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 contributed by Shubham Singh

Javascript




<script>
    // Javascript code to implement the above approach
     
    let n = 14, k = 2;
    document.write(Josephus(n, k));
     
    function Josephus(n, k)
    {
      k--;
      let arr = new Array(n);
      for (let i = 0; i < n; i++)
      {
       
          // Makes all the 'n' people alive by
        // assigning them value = 1
        arr[i] = 1;
      }
       
      // Cut = 0 gives the sword to 1st person.
      let cnt = 0, cut = 0,
      num = 1;
       
      // Loop continues till n-1 person dies.
      while (cnt < (n - 1))
      {
       
       // Checks next (kth) alive persons.
        while (num <= k)
        {
          cut++;
          cut = cut % n;
 
          // Checks and resolves overflow
          // of Index.
          if (arr[cut] == 1)
          {
           
               // Updates the number of persons
            // alive.
            num++;
          }
        }
         
        // refreshes value to 1 for next use.
        num = 1;
        arr[cut] = 0; // Kills the person at position of 'cut'
         
         // Updates the no. of killed persons.
        cnt++;
        cut++;
         
        // Checks and resolves overflow of Index.
        cut = cut % n;
         
        // Checks the next alive person the
        // sword is to be given.
        while (arr[cut] == 0)
        {
          cut++;
           
          // Checks and resolves overflow
          // of Index.
          cut = cut % n;
        }
      }
       
       // Output is the position of the last
      // man alive(Index + 1);
      return cut + 1;
    }
   
  // This code is contributed by decode2207.
</script>

Output

13

Time Complexity: O(N2)
Auxiliary Space: O(N)

Josephus Problem in Linear Time and Constant Space:

Follow the below steps to Solve the Problem (Approach):

  • Initialize variables i and ans with 1 and 0 respectively.
  • Run a while loop till i <= N:
    • Update ans with (ans + k) % i.
    • Increment i by 1.
  • Return ans + 1 as the required answer.

Below is the Implementation of the above Steps:

C++




// C++ code to Implement Josephus Problem
 
#include <iostream>
using namespace std;
 
int Josephus(int N, int k)
{
 
    // Initialize variables i and ans with 1 and 0
    // respectively.
 
    int i = 1, ans = 0;
 
    // Run a while loop till i <= N
 
    while (i <= N) {
 
        // Update the Value of ans and Increment i by 1
        ans = (ans + k) % i;
        i++;
    }
 
    // Return required answer
    return ans + 1;
}
 
// main function
int main()
{
 
    int N = 14, k = 2;
    cout << Josephus(N, k) << endl;
    return 0;
}
 
// This code is presented by Akash Mangal

C




// C Program to Implement Josephus Problem
 
#include <stdio.h>
 
int Josephus(int N, int k)
{
 
    // Initialize variables i and ans with 1 and 0
    // respectively.
 
    int i = 1, ans = 0;
 
    // Run a while loop till i <= N
 
    while (i <= N) {
 
        // Update the Value of ans and Increment i by 1
        ans = (ans + k) % i;
        i++;
    }
 
    // Return required answer
    return ans + 1;
}
 
// main function
int main()
{
 
    int N = 14, k = 2;
    printf("%d", Josephus(N, k));
    return 0;
}
 
// This code is presented by Akash Mangal

Output

13

Time Complexity: O(N)
Auxiliary Space: O(1)

Josephus Problem using Recursion:

Below is the idea to solve the problem:

The problem has the following recursive structure. josephus(n, k) = (josephus(n – 1, k) + k-1) % n + 1 and josephus(1, k) = 1

After the first person (kth from the beginning) is killed, n-1 persons are left. Make recursive call for 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 make adjustments to the position returned by Josephus(n – 1, k). 

Below is the Implementation of the above idea.

C++




// C++ code to implement the idea
 
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to implement the Josephus problem
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 code
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) the space used in recursion call stack

Please visit set-2: 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 if you want to share more information about the topic discussed above. 


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