Solving Josephus problem using Java

The Josephus problem (or Josephus permutation) is a theoretical problem related to a certain counting-out game.

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People are standing in a circle waiting to be executed. Counting begins at a specified point in the circle and proceeds around the circle in a specified direction. After a specified number of people are skipped, the next person is executed. The procedure is repeated with the remaining people, starting with the next person, going in the same direction and skipping the same number of people, until only one person remains, and is freed.

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Josephus problem

The source code example is given below

package com.test;

import java.util.List;

public class JosephProblem {

	public static void main(String[] args) {
		int winner = joseph(5, 3);
		System.out.println("winner is " + winner);
		winner = joseph(10, 3);
		System.out.println("winner is " + winner);
		winner = joseph(5, 2);
		System.out.println("winner is " + winner);
		winner = joseph(7, 3);
		System.out.println("winner is " + winner);

	public static int joseph(int noOfPeople, int remPosition) {
		int tempPos = remPosition - 1;
		int[] people = new int[noOfPeople];
		for (int i = 0; i < noOfPeople; i++) {
			people[i] = i + 1;

		int iteration = noOfPeople - 1;
		List<Integer> list = IntStream.of(people).boxed().collect(Collectors.toList());
		while (iteration > 0) {
			tempPos += remPosition - 1;
			if (tempPos > list.size() - 1) {
				tempPos = tempPos % list.size();
		return list.get(0);


Ananlysis of the logic

Let’s take an example of there are 5 people in a circle and execution starts clockwise at position 3, so we will find out the winner of the game.

So there are five people – 1 2 3 4 5 and execution starts clockwise at position 3. Therefore if we kill 3 then are left with the list something like – 1 2 4 5.

Now again we have to kill the person at position 3 clockwise counting from the last killed person. Now we need to execute the person at position 1. So executing the person at 1 we are left with the list 2 4 5. After executing another person we get list 2 4.

Now we are left with 2 persons in the list. So we have to calculate the position wisely to execute the person. Therefore the calculated position would be (starting position of the execution)%(no of people remaining in the list less than starting position) = 3%2 = 1. So finally the winner is 4.

The output comes after running the above code

winner is 4 winner is 4 winner is 3 winner is 4

Thanks for reading.

Soumitra Roy Sarkar

I am a professional Web developer, Enterprise Application developer, Software Engineer and Blogger. Connect me on Roy Tutorials Twitter Facebook  Google Plus Linkedin Or Email Me

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