# Minimum possible final health of the last monster in a game

• Difficulty Level : Medium
• Last Updated : 05 Aug, 2021

Given N monsters, each monster has initial health h[i] which is an integer. A monster is alive if its health is greater than 0. In each turn a random monster kills another random monster, the monster which is attacked, its health reduces by the amount of health of the attacking monster. This process is continued until a single monster is left. What will be the minimum possible health of the last remained monster. In others words, the task is to play the game in such a way that monster which is left in the end has the least possible health.
Examples:

Input: h[] = {2, 14, 28, 56}
Output:
When only the first monster keeps on attacking the remaining 3 monsters, the final health of the last monster will be 2, which is minimum.
Input: h[] = {7, 17, 9, 100, 25}
Output: 1
Input: h[] = {5, 5, 5}
Output:

Approach: It can be observed from the problem that one has to find a certain value of health of the monster, let’s say k which can kill other monsters including self. Once this crucial observation is made problem becomes easy. Suppose we have two monsters with health h1 and h2, and let’s say h2 > h1. We can see that in a random choice, the optimal way would be to pick a monster with lower health and reduce the health of the other monster till its health becomes less than the health of the attacking monster. After that we will pick the second monster whose health has became less than h1 and the process will continue till only one monster is left. So at last we will be left with the minimum value which would be gcd(h1, h2). This gcd method can be extended for all the monsters.
So our resultant minimum possible health of the monster will be the gcd of all the health of given monsters i.e. H(min) = gcd(h1, h2, …, hn).
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the gcd of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to return the minimum``// possible health for the monster``int` `solve(``int``* health, ``int` `n)``{``    ``// gcd of first and second element``    ``int` `currentgcd = gcd(health[0], health[1]);` `    ``// gcd for all subsequent elements``    ``for` `(``int` `i = 2; i < n; ++i) {``        ``currentgcd = gcd(currentgcd, health[i]);``    ``}``    ``return` `currentgcd;``}` `// Driver code``int` `main()``{``    ``int` `health[] = { 4, 6, 8, 12 };``    ``int` `n = ``sizeof``(health) / ``sizeof``(health[0]);``    ``cout << solve(health, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the gcd of two numbers``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == ``0``)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to return the minimum``// possible health for the monster``static` `int` `solve(``int` `health[], ``int` `n)``{``    ``// gcd of first and second element``    ``int` `currentgcd = gcd(health[``0``], health[``1``]);` `    ``// gcd for all subsequent elements``    ``for` `(``int` `i = ``2``; i < n; ++i)``    ``{``        ``currentgcd = gcd(currentgcd, health[i]);``    ``}``    ``return` `currentgcd;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `health[] = { ``4``, ``6``, ``8``, ``12` `};``    ``int` `n = health.length;``    ``System.out.println(solve(health, n));``}``}` `// This code is contributed by``// Surendra_Gangwar`

## Python3

 `# Python3 implementation of the approach` `# Function to return the gcd of two numbers``def` `gcd(a, b):` `    ``if` `(a ``=``=` `0``):``        ``return` `b``    ``return` `gcd(b ``%` `a, a)` `# Function to return the minimum``# possible health for the monster``def` `solve(health, n):``    ` `    ``# gcd of first and second element``    ``currentgcd ``=` `gcd(health[``0``], health[``1``])` `    ``# gcd for all subsequent elements``    ``for` `i ``in` `range``(``2``, n):``        ``currentgcd ``=` `gcd(currentgcd,``                         ``health[i])``    ``return` `currentgcd` `# Driver code``health ``=` `[``4``, ``6``, ``8``, ``12``]``n ``=` `len``(health)``print``(solve(health, n))` `# This code is contributed by mohit kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the gcd of two numbers``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(a == 0)``        ``return` `b;``    ``return` `gcd(b % a, a);``}` `// Function to return the minimum``// possible health for the monster``static` `int` `solve(``int` `[]health, ``int` `n)``{``    ``// gcd of first and second element``    ``int` `currentgcd = gcd(health[0], health[1]);` `    ``// gcd for all subsequent elements``    ``for` `(``int` `i = 2; i < n; ++i)``    ``{``        ``currentgcd = gcd(currentgcd, health[i]);``    ``}``    ``return` `currentgcd;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``int` `[]health = { 4, 6, 8, 12 };``    ``int` `n = health.Length;``    ``Console.WriteLine(solve(health, n));``}``}` `// This code is contributed by Arnab Kundu`

## PHP

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## Javascript

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Output:

`2`

Time Complexity: O(N * log(MAX)) where N is the size of the array and MAX is the maximum number in the array.
We are running a loop that takes O(N) time. Also, the GCD function takes O(log(min(A, B)), and in the worst case when A and B are the same and A = B = MAX then the GCD function takes O(log(MAX)) time. So, overall time complexity = O(N * log(MAX))
Auxiliary Space: O(log(MAX))

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