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Find the minimum possible health of the winning player

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  • Difficulty Level : Easy
  • Last Updated : 10 Jul, 2022
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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 <bits/stdc++.h>
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




<script>
 
 
// Javascript implementation of the above approach
 
// Function to return the minimum possible
// health of the last player
function minHealth(health, n)
{
 
    // Find the GCD of the array elements
    var gcd = 0;
    for (var i = 0; i < n; i++)
    {
        gcd = __gcd(gcd, health[i]);
    }
    return gcd;
}
 
function __gcd(a, b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
var health = [ 5, 6, 1, 2, 3, 4 ];
var n = health.length;
document.write(minHealth(health, n));
 
</script>

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.


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