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.

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<bits/stdc++.h> ` ` ` `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; ` `} ` |

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## 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. ` |

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## 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 ` |

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## 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. ` |

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## PHP

`<?php ` `// PHP program to calculate ` `// maximum number of groups ` ` ` `// Implements above mentioned steps. ` `function` `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` `%= 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. ` `?> ` |

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**Output :**

3

This article is contributed by **Anuj Chauhan(anuj0503)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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