Josephus Problem | (Iterative Solution)

There are N Children are seated on N chairs arranged around a circle. The chairs are numbered from 1 to N. The game starts going in circles counting the children starting with the first chair. Once the count reaches K, that child leaves the game, removing his/her chair. The game starts again, beginning with the next chair in the circle. The last child remaining in the circle is the winner. Find the child that wins the game.


Input : N = 5, K = 2
Output : 3
Firstly, the child at position 2 is out, 
then position 4 goes out, then position 1
Finally, the child at position 5 is out. 
So the position 3 survives.

Input : 7 4
Output : 2

We have discussed a recursive solution for Josephus Problem . The given solution is better than the recursive solution of Josephus Solution which is not suitable for large inputs as it gives stack overflow. The time complexity is O(N).

Approach – In the algorithm, we use sum variable to find out the chair to be removed. The current chair position is calculated by adding the chair count K to the previous position i.e. sum and modulus of the sum. At last we return sum+1 as numbering starts from 1 to N.





// Iterative solution for Josephus Problem 
#include <bits/stdc++.h>
using namespace std;
// Function for finding the winning child.
long long int find(long long int n, long long int k)
    long long int sum = 0, i;
    // For finding out the removed 
    // chairs in each iteration
    for (i = 2; i <= n; i++)
        sum = (sum + k) % i;
    return sum + 1;
// Driver function to find the winning child
int main()
    int n = 14, k = 2;
    cout << find(n, k);
    return 0;




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