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 codeint main(){ int x = 4, y = 5, z = 6; cout << maxItems(x, y, z); return 0;} |
Java
// Java implementation of the approachimport 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 codepublic 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 approachusing 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 codestatic 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 approachconst 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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