Maximum number of 3-person teams formed from two groups

Given two integers N1 and N2 where, N1 is the number of people in group 1 and N2 is the number of people in group 2. The task is to count the maximum number of 3-person teams that can be formed when at least a single person is chosen from both the groups.
Examples:

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 

Approach: Choose a single person from the team with less members and choose 2 persons from the team with more members (while possible) and update count = count + 1. Print the count in the end.
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 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 members from group 1
        // and a single member from group 2
        if (N1 > N2) {
            N1 -= 2;
            N2 -= 1;
        }

        // Choose 2 members from group 2
        // and a single member 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 members from group 1
            // and a single member from group 2
            if (N1 > N2) {
                N1 -= 2;
                N2 -= 1;
            }

            // Choose 2 members from group 2
            // and a single member 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 members from group 1
        # and a single member from group 2
        if (N1 > N2):
            N1 -= 2
            N2 -= 1


        # Choose 2 members from group 2
        # and a single member 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 members from group 1
            // and a single member from group 2
            if (N1 > N2) {
                N1 -= 2;
                N2 -= 1;
            }

            // Choose 2 members from group 2
            // and a single member 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

<?php
// PHP implementation of the approach

// Function to return the count
// of maximum teams possible
function maxTeams($N1, $N2)
{
    $count = 0 ;

    // While it is possible to form a team
    while ($N1 > 0 && $N2 > 0 &&
                $N1 + $N2 >= 3)
    {

        // Choose 2 members from group 1
        // and a single member from group 2
        if ($N1 > $N2)
        {
            $N1 -= 2;
            $N2 -= 1;
        }

        // Choose 2 members from group 2
        // and a single member 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
?>

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the count
    // of maximum teams possible
    function maxTeams(N1, N2)
    {

        let count = 0;

        // While it is possible to form a team
        while (N1 > 0 && N2 > 0 && N1 + N2 >= 3) {

            // Choose 2 members from group 1
            // and a single member from group 2
            if (N1 > N2) {
                N1 -= 2;
                N2 -= 1;
            }

            // Choose 2 members from group 2
            // and a single member from group 1
            else {
                N1 -= 1;
                N2 -= 2;
            }

            // Update the count
            count++;
        }

        // Return the count
        return count;
    }

    let N1 = 4, N2 = 5;
      document.write(maxTeams(N1, N2));

    // This code is contributed by divyeshrabadiya07.
</script>

Output:

3

Time Complexity: O(min(N1,N2))
Auxiliary Space: O(1)