Find last 2 survivors in N persons standing in a circle after killing next to immediate neighbour

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++

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// 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;
}

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Java

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// 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

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Output: 

1 2

 

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

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