Josephus problem | Set 1 (A O(n) Solution)
- Difficulty Level : Medium
- Last Updated : 07 Jun, 2022
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 person 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 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 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.
If n = 7 and k = 3, then the safe position is 4. The persons at positions 3, 6, 2, 7, 5, and 1 are killed in order, and the person at position 4 survives.

The problem has the following recursive structure.
josephus(n, k) = (josephus(n - 1, k) + k-1) % n + 1 josephus(1, k) = 1
After the first person (kth from the 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).
The following is a 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> |
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 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.
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); } |
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> |
13
Time Complexity: O(n2)
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 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.
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 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> |
13
Please visit set-2: Josephus problem | Set 2 (A Simple Solution when k = 2)
Source:
http://en.wikipedia.org/wiki/Josephus_problem
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