Skip to content
Related Articles

Related Articles

Find last 2 survivors in N persons standing in a circle after killing next to immediate neighbour
  • Last Updated : 28 Nov, 2020

Given an integer N representing N persons standing in a circle, the task is to find the last 2 persons remaining when a person kills their next to the immediate neighbor in a clockwise direction.

Examples:  

Input: N = 5 
Output: 1 4 
Explanation: 
Initially: 1 2 3 4 5 
=> 1 kills 3 
Standing: 1 2 4 5
=> 2 kills 5 
Standing: 1 2 4
=> 4 kills 2 
Final Standing: 1 4

Input: N = 2 
Output: 1 2 

Naive Approach: A simple approach is to keep a bool array of size N to keep track of whether a person is alive or not.  



  • Initially, the boolean array will be true for all persons.
  • Keep two pointers, one at the current alive person and second to store previous current person.
  • Once found a second alive neighbour from the current person, change its boolean value to false.
  • Then again current is updated to next alive from previous.
  • This process will continue till last two persons survive.

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

Efficient Approach: An efficient approach is to remove the person, if dead, from the data structure so that it is not traversed again.

  • After one complete round only, there will be only N/2 person, at max.
  • Then in the next round it will be left with N/4 person and so on until a number of alive people become 2.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Node for a Linked List
struct Node {
    int val;
    struct Node* next;
 
    Node(int _val)
    {
        val = _val;
        next = NULL;
    }
};
 
// Function to find the last 2 survivors
void getLastTwoPerson(int n)
{
    // Total is the count
    // of alive people
    int total = n;
    struct Node* head = new Node(1);
    struct Node* temp = head;
 
    // Initiating the list of n people
    for (int i = 2; i <= n; i++) {
        temp->next = new Node(i);
        temp = temp->next;
    }
 
    temp->next = head;
    temp = head;
 
    struct Node* del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2) {
 
        // Del represent next person
        // to be deleted or killed
        del = temp->next->next;
        temp->next->next
            = temp->next->next->next;
        temp = temp->next;
        free(del);
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    cout << temp->val << " "
         << temp->next->val;
}
 
// Driver code
int main()
{
    int n = 2;
 
    getLastTwoPerson(n);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG{
 
// Node for a Linked List
static class Node
{
    int val;
    Node next;
 
    Node(int _val)
    {
        val = _val;
        next = null;
    }
};
 
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
     
    // Total is the count
    // of alive people
    int total = n;
    Node head = new Node(1);
    Node temp = head;
 
    // Initiating the list of n people
    for(int i = 2; i <= n; i++)
    {
        temp.next = new Node(i);
        temp = temp.next;
    }
 
    temp.next = head;
    temp = head;
 
    Node del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2)
    {
 
        // Del represent next person
        // to be deleted or killed
        del = temp.next.next;
        temp.next.next = temp.next.next.next;
        temp = temp.next;
        del = null;
         
        System.gc();
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    System.out.print(temp.val + " " +
                     temp.next.val);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 2;
 
    getLastTwoPerson(n);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Node for a Linked List
class newNode:
     
    def __init__(self, val):
         
        self.val = val
        self.next = None
 
# Function to find the last 2 survivors
def getLastTwoPerson(n):
     
    # Total is the count
    # of alive people
    total = n
    head = newNode(1)
    temp = head
 
    # Initiating the list of n people
    for i in range(2, n + 1, 1):
        temp.next = newNode(i)
        temp = temp.next
 
    temp.next = head
    temp = head
 
    de = None
 
    # Total != 2 is terminating
    # condition because
    # at last only two-person
    # will remain alive
    while (total != 2):
         
        # de represent next person
        # to be deeted or killed
        de = temp.next.next
        temp.next.next = temp.next.next.next
        temp = temp.next
        del de
        total -= 1
 
    # Last two person to
    # survive (in any order)
    print(temp.val, temp.next.val)
 
# Driver code
if __name__ == '__main__':
     
    n = 2
     
    getLastTwoPerson(n)
 
# This code is contributed by SURENDRA_GANGWAR

C#




// C# implementation of the approach
using System;
 
class GFG{
 
// Node for a Linked List
class Node
{
    public int val;
    public Node next;
 
    public Node(int _val)
    {
        val = _val;
        next = null;
    }
};
 
// Function to find the last 2 survivors
static void getLastTwoPerson(int n)
{
     
    // Total is the count
    // of alive people
    int total = n;
    Node head = new Node(1);
    Node temp = head;
 
    // Initiating the list of n people
    for(int i = 2; i <= n; i++)
    {
        temp.next = new Node(i);
        temp = temp.next;
    }
 
    temp.next = head;
    temp = head;
 
    Node del;
 
    // Total != 2 is terminating
    // condition because
    // at last only two-person
    // will remain alive
    while (total != 2)
    {
 
        // Del represent next person
        // to be deleted or killed
        del = temp.next.next;
        temp.next.next = temp.next.next.next;
        temp = temp.next;
        del = null;
         
        total -= 1;
    }
 
    // Last two person to
    // survive (in any order)
    Console.Write(temp.val + " " +
                  temp.next.val);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 2;
 
    getLastTwoPerson(n);
}
}
 
// This code is contributed by Amit Katiyar
Output: 
1 2

 

Time Complexity: O(N*log N) 
Auxiliary Space: O(N)
 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :