Maximum number of teams that can be formed with given persons

Given two integers N and M which denote the number of persons of Type1 and Type2 respectively. The task is to find the maximum number of teams that can be formed with these two types of persons. A team can contain either 2 persons of Type1 and 1 person of Type2 or 1 person of Type1 and 2 persons of Type2.
Examples:

Input: N = 2, M = 6 
Output: 2 
(Type1, Type2, Type2) and (Type1, Type2, Type2) are the two possible teams.
Input: N = 4, M = 5 
Output: 3 

Approach: A greedy approach is to choose 2 persons from the group which has more members and 1 person from the group with lesser members and update the count of persons in each of the group accordingly. Termination condition will be when no more teams can be formed.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if it possible
// to form a team with the given n and m
bool canFormTeam(int n, int m)
{

    // 1 person of Type1 and 2 persons of Type2
    // can be chosen
    if (n >= 1 && m >= 2)
        return true;

    // 1 person of Type2 and 2 persons of Type1
    // can be chosen
    if (m >= 1 && n >= 2)
        return true;

    // Cannot from a team
    return false;
}

// Function to return the maximum number of teams
// that can be formed
int maxTeams(int n, int m)
{
    // To store the required count of teams formed
    int count = 0;

    while (canFormTeam(n, m)) {
        if (n > m) {

            // Choose 2 persons of Type1
            n -= 2;

            // And 1 person of Type2
            m -= 1;
        }
        else {

            // Choose 2 persons of Type2
            m -= 2;

            // And 1 person of Type1
            n -= 1;
        }

        // Another team has been formed
        count++;
    }

    return count;
}

// Driver code
int main()
{
    int n = 4, m = 5;
    cout << maxTeams(n, m);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function that returns true
    // if it possible to form a
    // team with the given n and m
    static boolean canFormTeam(int n, int m)
    {

        // 1 person of Type1 and 2 persons
        // of Type2 can be chosen
        if (n >= 1 && m >= 2)
            return true;

        // 1 person of Type2 and 2 persons
        // of Type1 can be chosen
        if (m >= 1 && n >= 2)
            return true;

        // Cannot from a team
        return false;
    }

    // Function to return the maximum
    // number of teams that can be formed
    static int maxTeams(int n, int m)
    {
        // To store the required count
        // of teams formed
        int count = 0;

        while (canFormTeam(n, m))
        {
            if (n > m)
            {

                // Choose 2 persons of Type1
                n -= 2;

                // And 1 person of Type2
                m -= 1;
            }
            else
            {

                // Choose 2 persons of Type2
                m -= 2;

                // And 1 person of Type1
                n -= 1;
            }

            // Another team has been formed
            count++;
        }
        return count;
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 4, m = 5;
        System.out.println(maxTeams(n, m));
    }
}

// This code is contributed by Ryuga

Python3

# Python 3 implementation of the approach

# Function that returns true if it possible
# to form a team with the given n and m
def canFormTeam(n, m):

    # 1 person of Type1 and 2 persons of Type2
    # can be chosen
    if (n >= 1 and m >= 2):
        return True

    # 1 person of Type2 and 2 persons of Type1
    # can be chosen
    if (m >= 1 and n >= 2):
        return True

    # Cannot from a team
    return False

# Function to return the maximum number of teams
# that can be formed
def maxTeams(n, m):
    # To store the required count of teams formed
    count = 0

    while (canFormTeam(n, m)):
        if (n > m):
            # Choose 2 persons of Type1
            n -= 2

            # And 1 person of Type2
            m -= 1

        else:
            # Choose 2 persons of Type2
            m -= 2

            # And 1 person of Type1
            n -= 1

        # Another team has been formed
        count += 1

    return count

# Driver code
if __name__ == '__main__':
    n = 4
    m = 5
    print(maxTeams(n, m))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function that returns true if it possible
// to form a team with the given n and m
static bool canFormTeam(int n, int m)
{

    // 1 person of Type1 and 2 persons
    //  of Type2 can be chosen
    if (n >= 1 && m >= 2)
        return true;

    // 1 person of Type2 and 2 persons
    // of Type1 can be chosen
    if (m >= 1 && n >= 2)
        return true;

    // Cannot from a team
    return false;
}

// Function to return the maximum
// number of teams that can be formed
static int maxTeams(int n, int m)
{
    // To store the required count
    // of teams formed
    int count = 0;

    while (canFormTeam(n, m))
    {
        if (n > m)
        {

            // Choose 2 persons of Type1
            n -= 2;

            // And 1 person of Type2
            m -= 1;
        }
        else
        {

            // Choose 2 persons of Type2
            m -= 2;

            // And 1 person of Type1
            n -= 1;
        }

        // Another team has been formed
        count++;
    }
    return count;
}

// Driver code
public static void Main()
{
    int n = 4, m = 5;
    Console.WriteLine(maxTeams(n, m));
}
}

// This code is contributed by
// tufan_gupta2000

PHP

<?php
// PHP implementation of the approach

// Function that returns true if it possible
// to form a team with the given n and m
function canFormTeam($n, $m)
{

    // 1 person of Type1 and 2 persons
    // of Type2 can be chosen
    if ($n >= 1 && $m >= 2)
        return true;

    // 1 person of Type2 and 2 persons
    // of Type1 can be chosen
    if ($m >= 1 && $n >= 2)
        return true;

    // Cannot from a team
    return false;
}

// Function to return the maximum number
// of teams that can be formed
function maxTeams($n, $m)
{

    // To store the required count
    // of teams formed
    $count = 0;

    while (canFormTeam($n, $m))
    {
        if ($n > $m)
        {

            // Choose 2 persons of Type1
            $n -= 2;

            // And 1 person of Type2
            $m -= 1;
        }
        else
        {

            // Choose 2 persons of Type2
            $m -= 2;

            // And 1 person of Type1
            $n -= 1;
        }

        // Another team has been formed
        $count++;
    }

    return $count;
}

// Driver code
$n = 4;
$m = 5;
echo maxTeams($n, $m);

// This code is contributed by mits
?>

Javascript

<script>
// javascript implementation of the approach
// Function that returns true

    // if it possible to form a
    // team with the given n and m
    function canFormTeam(n, m)
    {

        // 1 person of Type1 and 2 persons
        // of Type2 can be chosen
        if (n >= 1 && m >= 2)
            return true;

        // 1 person of Type2 and 2 persons
        // of Type1 can be chosen
        if (m >= 1 && n >= 2)
            return true;

        // Cannot from a team
        return false;
    }

    // Function to return the maximum
    // number of teams that can be formed
    function maxTeams(n , m)
    {

        // To store the required count
        // of teams formed
        var count = 0;

        while (canFormTeam(n, m))
        {
            if (n > m)
            {

                // Choose 2 persons of Type1
                n -= 2;

                // And 1 person of Type2
                m -= 1;
            }
            else
            {

                // Choose 2 persons of Type2
                m -= 2;

                // And 1 person of Type1
                n -= 1;
            }

            // Another team has been formed
            count++;
        }
        return count;
    }

    // Driver code
    var n = 4, m = 5;
    document.write(maxTeams(n, m));

// This code is contributed by todaysgaurav
</script>

Output:

3

Time Complexity: O(min(n, m))

Auxiliary Space: O(1)