Maximum items that can be bought with the given type of coins

Given three integers X, Y and Z which represent the number of coins to buy some items. The cost of items is given below: 

Item type	Cost
1	3 X coins
2	3 Y coins
3	3 Z coins
4	1 X coin + 1 Y coin + 1 Z coin

The task is to find the maximum number of items that can be bought with the given number of coins.

Input: X = 4, Y = 5, Z = 6 
Output: 4 
Buy 1 item of type 1: X = 1, Y = 5, Z = 6 
Buy 1 item of type 2: X = 1, Y = 2, Z = 6 
Buy 2 items of type 3: X = 1, Y = 2, Z = 0 
Total items bought = 1 + 1 + 2 = 4
Input: X = 6, Y = 7, Z = 9 
Output: 7 

Approach: The count of items of type1, type2 and type3 that can be bought will be X / 3, Y / 3 and Z / 3 respectively. Now, the number of coins will get reduced after buying these items as X = X % 3, Y = Y % 3 and Z = Z % 3. Since, buying the item of type 4 requires a coin from each of the type. So, the total items of type 4 that can be bought will be the minimum of X, Y and Z and the result will be the sum of these items which were bought from each of the type.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

const int COST = 3;

// Function to find maximum fruits
// Can buy from given values of x, y, z.
int maxItems(int x, int y, int z)
{

    // Items of type 1 that can be bought
    int type1 = x / COST;

    // Update the coins
    x %= COST;

    // Items of type 2 that can be bought
    int type2 = y / COST;

    // Update the coins
    y %= COST;

    // Items of type 3 that can be bought
    int type3 = z / COST;

    // Update the coins
    z %= COST;

    // Items of type 4 that can be bought
    // To buy a type 4 item, a coin
    // of each type is required
    int type4 = min(x, min(y, z));

    // Total items that can be bought
    int maxItems = type1 + type2 + type3 + type4;
    return maxItems;
}

// Driver code
int main()
{
    int x = 4, y = 5, z = 6;

    cout << maxItems(x, y, z);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

class GFG
{
static int COST = 3;

// Function to find maximum fruits
// Can buy from given values of x, y, z.
static int maxItems(int x, int y, int z)
{

    // Items of type 1 that can be bought
    int type1 = x / COST;

    // Update the coins
    x %= COST;

    // Items of type 2 that can be bought
    int type2 = y / COST;

    // Update the coins
    y %= COST;

    // Items of type 3 that can be bought
    int type3 = z / COST;

    // Update the coins
    z %= COST;

    // Items of type 4 that can be bought
    // To buy a type 4 item, a coin
    // of each type is required
    int type4 = Math.min(x, Math.min(y, z));

    // Total items that can be bought
    int maxItems = type1 + type2 + type3 + type4;
    return maxItems;
}

// Driver code
public static void main (String[] args)
{
    int x = 4, y = 5, z = 6;
    System.out.println(maxItems(x, y, z));
}
}

// This code is contributed by @tushil

Python3

# Python3 implementation of the approach
COST = 3;

# Function to find maximum fruits
# Can buy from given values of x, y, z.
def maxItems(x, y, z) :

    # Items of type 1 that can be bought
    type1 = x // COST;

    # Update the coins
    x %= COST;

    # Items of type 2 that can be bought
    type2 = y // COST;

    # Update the coins
    y %= COST;

    # Items of type 3 that can be bought
    type3 = z // COST;

    # Update the coins
    z %= COST;

    # Items of type 4 that can be bought
    # To buy a type 4 item, a coin
    # of each type is required
    type4 = min(x, min(y, z));

    # Total items that can be bought
    maxItems = type1 + type2 + type3 + type4;
    return maxItems;

# Driver code
if __name__ == "__main__" :

    x = 4; y = 5; z = 6;

    print(maxItems(x, y, z));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{
static int COST = 3;

// Function to find maximum fruits
// Can buy from given values of x, y, z.
static int maxItems(int x, int y, int z)
{

    // Items of type 1 that can be bought
    int type1 = x / COST;

    // Update the coins
    x %= COST;

    // Items of type 2 that can be bought
    int type2 = y / COST;

    // Update the coins
    y %= COST;

    // Items of type 3 that can be bought
    int type3 = z / COST;

    // Update the coins
    z %= COST;

    // Items of type 4 that can be bought
    // To buy a type 4 item, a coin
    // of each type is required
    int type4 = Math.Min(x, Math.Min(y, z));

    // Total items that can be bought
    int maxItems = type1 + type2 + type3 + type4;
    return maxItems;
}

// Driver code
static public void Main ()
{
    int x = 4, y = 5, z = 6;

    Console.Write (maxItems(x, y, z));
}
}

// This code is contributed by ajit..

Javascript

<script>

// Javascript implementation of the approach

const COST = 3;

// Function to find maximum fruits
// Can buy from given values of x, y, z.
function maxItems(x, y, z)
{

    // Items of type 1 that can be bought
    let type1 = parseInt(x / COST);

    // Update the coins
    x %= COST;

    // Items of type 2 that can be bought
    let type2 = parseInt(y / COST);

    // Update the coins
    y %= COST;

    // Items of type 3 that can be bought
    let type3 = parseInt(z / COST);

    // Update the coins
    z %= COST;

    // Items of type 4 that can be bought
    // To buy a type 4 item, a coin
    // of each type is required
    let type4 = Math.min(x, Math.min(y, z));

    // Total items that can be bought
    let maxItems = type1 + type2 + type3 + type4;
    return maxItems;
}

// Driver code
    let x = 4, y = 5, z = 6;

    document.write(maxItems(x, y, z));

</script>

Output:

4

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