# 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.

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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;` `}` |

## 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

`<?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.` `?>` |

## Javascript

`<script>` `// JavaScript program to find Cullen number` `// 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;` ` ` ` ` `let ans = (m + n) / 3;` ` ` `m %= 3;` ` ` `n %= 3;` ` ` ` ` `if` `(m > 0 && n > 0 && (m + n) >= 3)` ` ` `ans++;` ` ` ` ` `return` `ans;` `}` `// Driver Code` ` ` `let n = 4, m = 5;` ` ` `document.write(maxGroup(n, m));` `// This code is contributed by chinmoy1997pal.` `</script>` |

**Output :**

3

**Time Complexity:** O(1)**Auxiliary Space: **O(1)

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

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