# Maximum number of groups of size 3 containing two type of items

Given n instance of item A and m instance of item B. Find the maximum number of groups of size 3 that can be formed using these items such that all groups contain items of both types, i.e., a group should not have either all items of type A or all items of type B.

Total number of items of type A in the formed groups must be less than or equal to n.
Total number of items of type B in the formed groups must be less than or equal to m.

Examples :

```Input : n = 2 and m = 6.
Output : 2
In group 1, one item of type A and two items of
type B. Similarly, in the group 2, one item of
type A and two items of type B.
We have used 2 (<= n) items of type A and 4 (<= m)
items of type B.

Input : n = 4 and m = 5.
Output : 3
In group 1, one item of type A and two items of type B.
In group 2, one item of type B and two items of type A.
In group 3, one item of type A and two items of type B.
We have used 4 (<= n) items of type A and 5 (<= 5)
items of type B.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Observation:
1. There will be n groups possible if m >= 2n. Or there will be m groups possible, if n >= 2m.
2. Suppose n = 3 and m = 3, so one instance of item A will make a group with the two instance of item B and one instance of item B will make a group with the two instance of item A. So, maximum two groups are possible. So find the total number of such conditions with given n and m by dividing m and m by 3. After this, there can be 0, 1, 2 instances of each type can be left. For finding the number of groups for the left instances:
a) If n = 0 or m = 0, 0 group is possible.
b) If n + m >= 3, only 1 group is possible.

Algorithm for solving this problem:
1. If n >= 2m, maximum number of groups = n.
2. Else if m >= 2n, maximum number of groups = m.
3. Else If (m + n) % 3 == 0, maximum number of group = (m + n)/3;
4. Else maximum number of group = (n + m)/3. And set n = n%3 and m = m%3.
a) If n != 0 && m != 0 && (n + m) >= 3, add one to maximum number of groups.

Below is implementation of the above idea :

## C++

 `// C++ program to calculate  ` `// maximum number of groups ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// Implements above mentioned steps. ` `int` `maxGroup(``int` `n, ``int` `m) ` `{ ` `    ``if` `(n >= 2 * m) ` `        ``return` `n; ` `    ``if` `(m >= 2 * n) ` `        ``return` `m; ` `    ``if` `((m + n) % 3 == 0) ` `        ``return` `(m + n)/3; ` ` `  `    ``int` `ans = (m + n)/3; ` `    ``m %= 3; ` `    ``n %= 3; ` ` `  `    ``if` `(m && n && (m + n) >= 3) ` `        ``ans++; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 4, m = 5; ` `    ``cout << maxGroup(n, m) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to calculate  ` `// maximum number of groups ` `import` `java.io.*; ` ` `  `public` `class` `GFG{ ` `     `  `// Implements above mentioned steps. ` `static` `int` `maxGroup(``int` `n, ``int` `m) ` `{ ` `    ``if` `(n >= ``2` `* m) ` `        ``return` `n; ` `    ``if` `(m >= ``2` `* n) ` `        ``return` `m; ` `    ``if` `((m + n) % ``3` `== ``0``) ` `        ``return` `(m + n) / ``3``; ` ` `  `    ``int` `ans = (m + n) / ``3``; ` `    ``m %= ``3``; ` `    ``n %= ``3``; ` ` `  `    ``if` `(m > ``0` `&& n > ``0` `&& (m + n) >= ``3``) ` `        ``ans++; ` ` `  `    ``return` `ans; ` `} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `            ``int` `n = ``4``, m = ``5``; ` `    ``System.out.println(maxGroup(n, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python3 program to calculate maximum  ` `# number of groups ` ` `  `# Implements above mentioned steps ` `def` `maxGroup(n, m): ` `     `  `    ``if` `n >``=` `2` `*` `m: ` `        ``return` `n ` `    ``if` `m >``=` `2` `*` `n: ` `        ``return` `m ` `    ``if` `(m ``+` `n) ``%` `3` `=``=` `0``: ` `        ``return` `(m ``+` `n) ``/``/` `3` `    ``ans ``=` `(m ``+` `n) ``/``/` `3` `    ``m ``=` `m ``%` `3` `    ``n ``=` `n ``%` `3` `     `  `    ``if` `m ``and` `n ``and` `(m ``+` `n) >``=` `3``: ` `        ``ans ``+``=` `1` `    ``return` `ans  ` `     `  `# Driver Code ` `n, m ``=` `4``, ``5` `print``(maxGroup(n, m)) ` ` `  `# This code is contributed ` `# by Mohit kumar 29 `

## C#

 `// C# program to calculate  ` `// maximum number of groups ` `using` `System; ` ` `  `public` `class` `GFG{ ` `     `  `// Implements above mentioned steps. ` `static` `int` `maxGroup(``int` `n, ``int` `m) ` `{ ` `    ``if` `(n >= 2 * m) ` `        ``return` `n; ` `    ``if` `(m >= 2 * n) ` `        ``return` `m; ` `    ``if` `((m + n) % 3 == 0) ` `        ``return` `(m + n) / 3; ` ` `  `    ``int` `ans = (m + n) / 3; ` `    ``m %= 3; ` `    ``n %= 3; ` ` `  `    ``if` `(m > 0 && n > 0 && (m + n) >= 3) ` `        ``ans++; ` ` `  `    ``return` `ans; ` `} ` ` `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `n = 4, m = 5; ` `        ``Console.WriteLine(maxGroup(n, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `= 2 * ``\$m``) ` `        ``return` `n; ` `    ``if` `(``\$m` `>= 2 * ``\$n``) ` `        ``return` `m; ` `    ``if` `(((``\$m` `+ ``\$n``) % 3) == 0) ` `        ``return` `(``\$m` `+ ``\$n``) / 3; ` ` `  `    ``\$ans` `= (``\$m` `+ ``\$n``) / 3; ` `    ``\$m` `%= 3; ` `    ``\$n` `%= 3; ` ` `  `    ``if` `(``\$m` `&& ``\$n` `&& (``\$m` `+ ``\$n``) >= 3) ` `        ``\$ans``++; ` ` `  `    ``return` `\$ans``; ` `} ` ` `  `// Driver code ` `\$n` `= 4; ``\$m` `= 5; ` `echo` `maxGroup(``\$n``, ``\$m``) ; ` ` `  `// This code is contributed  ` `// by nitin mittal.  ` `?> `

Output :

```3
```

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