# 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:
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:

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 ``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`

## Javascript

 ``

## PHP

 ` 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``?>`

Output
```3

```

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

New Approach:- Here, another approach we can calculate the maximum number of teams that can be formed by dividing the total number of people from both groups by 3, rounding down to the nearest integer. This is because each team consists of 3 people.

However, we need to make sure that at least one person is chosen from both groups. So, if the number of people in one of the groups is less than 3, we cannot form a team and the answer would be 0. Otherwise, the answer would be the maximum number of teams that can be formed as calculated earlier.

Algorithm:-

• Calculate the total number of people by adding N1 and N2.
• Check if N1, N2, or the total number of people is less than 3. If so, it is impossible to form a team and the function returns 0.
• Calculate the maximum number of teams possible by dividing the total number of people by 3 (since each team must have 3 members).
• Return the maximum number of teams as the output of the function.
• In the driver code, the function is called with the values N1=4 and N2=5, and the result is printed.

Here’s the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to return the count of maximum teams possible``int` `maxTeams(``int` `N1, ``int` `N2) {``    ``int` `total_people = N1 + N2;``    ``if` `(N1 < 1 || N2 < 1 || total_people < 3) {``        ``// If either N1 or N2 is less than 1, or the total number of people``        ``// is less than 3, it's not possible to form a team.``        ``return` `0;``    ``}` `    ``// Calculate the maximum number of teams that can be formed by dividing``    ``// the total number of people by 3 (each team requires 3 people).``    ``int` `max_teams = total_people / 3;``    ``return` `max_teams;``}` `// Driver code``int` `main() {``    ``int` `N1 = 4;``    ``int` `N2 = 5;``    ``cout << maxTeams(N1, N2) << endl;``    ``return` `0;``}`

## Java

 `public` `class` `Main {``    ``// Function to return the count of maximum teams``    ``// possible``    ``static` `int` `maxTeams(``int` `N1, ``int` `N2)``    ``{``        ``int` `totalPeople = N1 + N2;``        ``if` `(N1 < ``1` `|| N2 < ``1` `|| totalPeople < ``3``) {``            ``// If either N1 or N2 is less than 1, or the``            ``// total number of people is less than 3, it's``            ``// not possible to form a team.``            ``return` `0``;``        ``}` `        ``// Calculate the maximum number of teams that can be``        ``// formed by dividing the total number of people by``        ``// 3 (each team requires 3 people).``        ``int` `maxTeams = totalPeople / ``3``;``        ``return` `maxTeams;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `N1 = ``4``;``        ``int` `N2 = ``5``;``        ``System.out.println(maxTeams(N1, N2));``    ``}``}`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count of maximum teams possible`  `def` `maxTeams(N1, N2):``    ``total_people ``=` `N1 ``+` `N2``    ``if` `N1 < ``1` `or` `N2 < ``1` `or` `total_people < ``3``:``        ``# Cannot form a team with less than 3 people or``        ``# if at least one group has no people``        ``return` `0``    ``max_teams ``=` `total_people ``/``/` `3``    ``return` `max_teams`  `# Driver code``N1 ``=` `4``N2 ``=` `5``print``(maxTeams(N1, N2))`

## C#

 `using` `System;` `public` `class` `GFG``{``    ``// Function to return the count of maximum teams possible``    ``public` `static` `int` `MaxTeams(``int` `N1, ``int` `N2)``    ``{``        ``int` `totalPeople = N1 + N2;``        ``if` `(N1 < 1 || N2 < 1 || totalPeople < 3)``        ``{``            ``// If either N1 or N2 is less than 1, or the total number of people``            ``// is less than 3, it's not possible to form a team.``            ``return` `0;``        ``}` `        ``// Calculate the maximum number of teams that can be formed by dividing``        ``// the total number of people by 3 (each team requires 3 people).``        ``int` `maxTeams = totalPeople / 3;``        ``return` `maxTeams;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `N1 = 4;``        ``int` `N2 = 5;``        ``Console.WriteLine(MaxTeams(N1, N2));``    ``}``}`

## Javascript

 `// Function to return the count of maximum teams possible``function` `maxTeams(N1, N2) {``    ``const totalPeople = N1 + N2;``    ` `    ``// Check if either N1 or N2 is less than 1, or the total number of people``    ``// is less than 3, it's not possible to form a team.``    ``if` `(N1 < 1 || N2 < 1 || totalPeople < 3) {``        ``return` `0;``    ``}` `    ``// Calculate the maximum number of teams that can be formed by dividing``    ``// the total number of people by 3 (each team requires 3 people).``    ``const maxTeamsCount = Math.floor(totalPeople / 3);``    ``return` `maxTeamsCount;``}` `// Driver code``const N1 = 4;``const N2 = 5;``console.log(maxTeams(N1, N2)); ``// Output: 3`

Output:-

`3`

Time Complexity:- The time complexity of this code is O(1)
Auxiliary Space:- The auxiliary space complexity of this code is also O(1)

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