# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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.. `

Output:

```4
```

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