# Find the minimum possible health of the winning player

• Difficulty Level : Easy
• Last Updated : 10 Jul, 2022

Given an array health[] where health[i] is the health of the ith player in a game, any player can attack any other player in the game. The health of the player being attacked will be reduced by the amount of health the attacking player has. The task is to find the minimum possible health of the winning player.
Examples:

Input: health[] = {4, 6, 8}
Output:
4 attacks 6, health[] = {4, 2, 8}
2 attacks 4 twice, health[] = {0, 2, 8}
2 attacks 8 four times, health[] = {0, 2, 0}
Input: health[] = {4, 1, 5, 3}
Output:

Approach: In order to minimize the health of the last player, only the player with the smaller health will attack a player with the larger health and by doing so if only two players are involved then the minimum health of the last player is nothing but the GCD of the initial healths of the two players. So, the result will be the GCD of all the elements of the given array.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum possible``// health of the last player``int` `minHealth(``int` `health[], ``int` `n)``{` `    ``// Find the GCD of the array elements``    ``int` `gcd = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``gcd = __gcd(gcd, health[i]);``    ``}` `    ``return` `gcd;``}` `// Driver code``int` `main()``{``    ``int` `health[] = { 5, 6, 1, 2, 3, 4 };``    ``int` `n = ``sizeof``(health) / ``sizeof``(``int``);` `    ``cout << minHealth(health, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{` `// Function to return the minimum possible``// health of the last player``static` `int` `minHealth(``int` `health[], ``int` `n)``{` `    ``// Find the GCD of the array elements``    ``int` `gcd = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``gcd = __gcd(gcd, health[i]);``    ``}``    ``return` `gcd;``}` `static` `int` `__gcd(``int` `a, ``int` `b)``{``    ``return` `b == ``0` `? a : __gcd(b, a % b);    ``}` `// Driver code``public` `static` `void` `main(String []args)``{``    ``int` `health[] = { ``5``, ``6``, ``1``, ``2``, ``3``, ``4` `};``    ``int` `n = health.length;` `    ``System.out.println(minHealth(health, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach``from` `math ``import` `gcd` `# Function to return the minimum possible``# health of the last player``def` `minHealth(health, n) :` `    ``# Find the GCD of the array elements``    ``__gcd ``=` `0``;``    ` `    ``for` `i ``in` `range``(n) :``        ``__gcd ``=` `gcd(__gcd, health[i]);` `    ``return` `__gcd;` `# Driver code``if` `__name__ ``=``=` `"__main__"` `:` `    ``health ``=` `[ ``5``, ``6``, ``1``, ``2``, ``3``, ``4` `];``    ``n ``=` `len``(health);` `    ``print``(minHealth(health, n));` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the above approach``using` `System;``    ` `class` `GFG``{` `// Function to return the minimum possible``// health of the last player``static` `int` `minHealth(``int` `[]health, ``int` `n)``{` `    ``// Find the GCD of the array elements``    ``int` `gcd = 0;``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``gcd = __gcd(gcd, health[i]);``    ``}``    ``return` `gcd;``}` `static` `int` `__gcd(``int` `a, ``int` `b)``{``    ``return` `b == 0 ? a : __gcd(b, a % b);    ``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]health = { 5, 6, 1, 2, 3, 4 };``    ``int` `n = health.Length;` `    ``Console.WriteLine(minHealth(health, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`1`

Time Complexity : O(log(a+b)) ,where a and b are elements of health array.

Space Complexity : O(1) ,as we are not using any extra space.

My Personal Notes arrow_drop_up