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.

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++

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// 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

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// 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

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# 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#

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// 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

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<?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

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