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

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  • Last Updated : 03 Mar, 2022

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 typeCost
13 X coins
23 Y coins
33 Z coins
41 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:
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:
 

 

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)


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