# 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++

 `// C++ implementation of the approach`   `#include ` `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 deleted 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`

## Javascript

 ``

Output:

`1 2`

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

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