Máximo de artículos que se pueden comprar con el tipo de monedas dado

Dados tres números enteros X , Y y Z que representan el número de monedas para comprar algunos artículos. El costo de los artículos se da a continuación: 
 

La tarea es encontrar la cantidad máxima de artículos que se pueden comprar con la cantidad dada de monedas.
 

Entrada: X = 4, Y = 5, Z = 6 
Salida: 4 
Comprar 1 artículo de tipo 1: X = 1, Y = 5, Z = 6 
Comprar 1 artículo de tipo 2: X = 1, Y = 2, Z = 6 
Compra 2 artículos de tipo 3: X = 1, Y = 2, Z = 0 
Total de artículos comprados = 1 + 1 + 2 = 4
Entrada: X = 6, Y = 7, Z = 9 
Salida: 7 
 

Enfoque: El recuento de artículos de tipo1 , tipo2 y tipo3 que se pueden comprar será X/3 , Y/3 y Z/3 respectivamente. Ahora, la cantidad de monedas se reducirá después de comprar estos artículos como X = X % 3 , Y = Y % 3 y Z = Z % 3 . Dado que comprar el artículo del tipo 4 requiere una moneda de cada tipo. Entonces, el total de artículos del tipo 4 que se pueden comprar será el mínimo de X , Y y Z y el resultado será la suma de estos artículos que se compraron de cada tipo.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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 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# 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..

<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>

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