两组最多可容纳3人的团队

给定两个整数N1和N2 ，其中N1是第1组的人数， N2是第2组的人数。任务是计算至少一个以上时可以组建的3人团队的最大数量。从两个组中都选择一个人。

例子：

Input: N1 = 2, N2 = 8
Output: 2
Team 1: 2 members from group 2 and 1 member from group 1
Update: N1 = 1, N2 = 6
Team 2: 2 members from group 2 and 1 member from group 1
Update: N1 = 0, N2 = 4
No further teams can be formed.

Input: N1 = 4, N2 = 5
Output: 3

方法：从成员较少的团队中选择一个人，然后从成员较多的团队中选择2个人(如果可能)，并更新count = count + 1 。最后打印计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count
// of maximum teams possible
int maxTeams(int N1, int N2)
{
  
    int count = 0;
  
    // While it is possible to form a team
    while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) {
  
        // Choose 2 memebers from group 1
        // and a single memeber from group 2
        if (N1 > N2) {
            N1 -= 2;
            N2 -= 1;
        }
  
        // Choose 2 memebers from group 2
        // and a single memeber from group 1
        else {
            N1 -= 1;
            N2 -= 2;
        }
  
        // Update the count
        count++;
    }
  
    // Return the count
    return count;
}
  
// Driver code
int main()
{
  
    int N1 = 4, N2 = 5;
    cout << maxTeams(N1, N2);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG
{
    // Function to return the count
    // of maximum teams possible
    static int maxTeams(int N1, int N2)
    {
      
        int count = 0;
      
        // While it is possible to form a team
        while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) {
      
            // Choose 2 memebers from group 1
            // and a single memeber from group 2
            if (N1 > N2) {
                N1 -= 2;
                N2 -= 1;
            }
      
            // Choose 2 memebers from group 2
            // and a single memeber from group 1
            else {
                N1 -= 1;
                N2 -= 2;
            }
      
            // Update the count
            count++;
        }
      
        // Return the count
        return count;
    }
      
    // Driver code
    public static void main(String []args)
    {
      
        int N1 = 4, N2 = 5;
        System.out.println(maxTeams(N1, N2));
      
          
    }
  
}
  
// This code is contributed by ihritik

Python3

# Python3 implementation of the approach
  
  
# Function to return the count
# of maximum teams possible
def maxTeams(N1, N2):
  
  
    count = 0
  
    # While it is possible to form a team
    while (N1 > 0 and N2 > 0 and N1 + N2 >= 3) :
  
        # Choose 2 memebers from group 1
        # and a single memeber from group 2
        if (N1 > N2): 
            N1 -= 2
            N2 -= 1
          
  
        # Choose 2 memebers from group 2
        # and a single memeber from group 1
        else:
            N1 -= 1
            N2 -= 2
          
  
        # Update the count
        count=count+1
      
  
    # Return the count
    return count
  
      
# Driver code
N1 = 4
N2 = 5
print(maxTeams(N1, N2))
  
# This code is contributed by ihritik

C#

// C# implementation of the approach
  
using System;
class GFG
{
    // Function to return the count
    // of maximum teams possible
    static int maxTeams(int N1, int N2)
    {
      
        int count = 0;
      
        // While it is possible to form a team
        while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) {
      
            // Choose 2 memebers from group 1
            // and a single memeber from group 2
            if (N1 > N2) {
                N1 -= 2;
                N2 -= 1;
            }
      
            // Choose 2 memebers from group 2
            // and a single memeber from group 1
            else {
                N1 -= 1;
                N2 -= 2;
            }
      
            // Update the count
            count++;
        }
      
        // Return the count
        return count;
    }
      
    // Driver code
    public static void Main()
    {
      
        int N1 = 4, N2 = 5;
        Console.WriteLine(maxTeams(N1, N2));
      
          
    }
  
}
  
// This code is contributed by ihritik

PHP

 0 && $N2 > 0 &&
                $N1 + $N2 >= 3)
    { 
  
        // Choose 2 memebers from group 1 
        // and a single memeber from group 2 
        if ($N1 > $N2) 
        { 
            $N1 -= 2; 
            $N2 -= 1; 
        } 
  
        // Choose 2 memebers from group 2 
        // and a single memeber from group 1 
        else
        { 
            $N1 -= 1; 
            $N2 -= 2; 
        } 
  
        // Update the count 
        $count++; 
    } 
  
    // Return the count 
    return $count; 
} 
  
// Driver code 
$N1 = 4 ;
$N2 = 5 ;
  
echo maxTeams($N1, $N2);
  
// This code is contributed by Ryuga
?>

输出：

3

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