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Maximum number of groups of size 3 containing two type of items
  • Last Updated : 24 Mar, 2021

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;
}

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 contribued by chinmoy1997pal.
</script>

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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