# 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 `

Output:

```1 2
```

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

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