# Minimize cost to convert given two integers to zero using given operations

Given two integers X and Y, and two values cost1 and cost2, the task is to convert the given two numbers equal to zero at minimal cost by performing the following two types of operations:

• Increase or decrease any one of them by 1 at cost1.
• Increase or decrease both of them by 1 at cost2.

Examples:

Input: X = 1, Y = 3, cost1 = 391, cost2 = 555
Output: 1337
Explanation:
Reduce Y to 1 using the first operation twice and convert both X and Y from 1 to 0 using the second operation.
Hence, the total cost = 391 * 2 + 555 = 1337.

Input: X = 12, Y = 7, cost1 = 12, cost2 = 7
Output: 4
Explanation:
Reduce X to 7 using first operation and then convert both X and Y to 0 using the second operation.
Hence, the total cost = 12 * 5 + 7 * 7 = 109

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

Approach:
The most optimal way to solve the problem is:

• Reduce the maximum of X and Y to the minimum by using first operation. This increases the cost by abs(X – Y) * cost1.
• Then, reduce both X and Y to 0 using the second operation. This increase the cost by minimum of (X, Y) * cost2.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the minimum ` `// cost to make the two integers equal ` `// to zero using given operations ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to find out the minimum cost to  ` `// make two number X and Y equal to zero ` `int` `makeZero(``int` `x, ``int` `y, ``int` `a, ``int` `b) ` `{ ` `     `  `    ``// If x is greater than y then swap ` `    ``if``(x > y) ` `       ``x = y,  ` `       ``y = x; ` `     `  `    ``// Cost of making y equal to x  ` `    ``int` `tot_cost = (y - x) * a; ` ` `  `    ``// Cost if we choose 1st operation ` `    ``int` `cost1 = 2 * x * a; ` `     `  `    ``// Cost if we choose 2nd operation ` `    ``int` `cost2 = x * b; ` `     `  `    ``// Total cost ` `    ``tot_cost += min(cost1, cost2); ` `     `  `    ``cout << tot_cost; ` `} ` ` `  `// Driver code  ` `int` `main()  ` `{  ` `    ``int` `X = 1, Y = 3; ` `    ``int` `cost1 = 391, cost2 = 555; ` ` `  `    ``makeZero(X, Y, cost1, cost2); ` `}  ` ` `  `// This code is contributed by coder001 `

## Java

 `// Java implementation to find the minimum ` `// cost to make the two integers equal ` `// to zero using given operations ` `import` `java.util.*; ` `class` `GFG{ ` `     `  `// Function to find out the minimum cost to  ` `// make two number X and Y equal to zero ` `static` `void` `makeZero(``int` `x, ``int` `y, ``int` `a, ``int` `b) ` `{ ` `     `  `    ``// If x is greater than y then swap ` `    ``if``(x > y) ` `    ``{ ` `        ``int` `temp = x; ` `        ``x = y; ` `        ``y = temp; ` `    ``} ` `     `  `    ``// Cost of making y equal to x  ` `    ``int` `tot_cost = (y - x) * a; ` ` `  `    ``// Cost if we choose 1st operation ` `    ``int` `cost1 = ``2` `* x * a; ` `     `  `    ``// Cost if we choose 2nd operation ` `    ``int` `cost2 = x * b; ` `     `  `    ``// Total cost ` `    ``tot_cost += Math.min(cost1, cost2); ` `     `  `    ``System.out.print(tot_cost); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[])  ` `{  ` `    ``int` `X = ``1``, Y = ``3``; ` `    ``int` `cost1 = ``391``, cost2 = ``555``; ` ` `  `    ``makeZero(X, Y, cost1, cost2); ` `}  ` `} ` ` `  `// This code is contributed by Code_Mech `

## Python3

 `# Python3 implementation to find the ` `# minimum cost to make the two integers ` `# equal to zero using given operations ` ` `  ` `  `# Function to find out the minimum cost to make ` `# two number X and Y equal to zero ` `def` `makeZero(x, y, a, b): ` `     `  `    ``# If x is greater than y then swap ` `    ``if``(x > y): ` `        ``x, y ``=` `y, x ` `     `  `    ``# Cost of making y equal to x  ` `    ``tot_cost ``=` `(y ``-` `x) ``*` `a ` ` `  `    ``# Cost if we choose 1st operation ` `    ``cost1 ``=` `2` `*` `x ``*` `a ` `     `  `    ``# Cost if we choose 2nd operation ` `    ``cost2 ``=` `x ``*` `b ` `     `  `    ``# Total cost ` `    ``tot_cost``+``=` `min``(cost1, cost2) ` `     `  `    ``print``(tot_cost) ` `     `  ` `  `if` `__name__ ``=``=``"__main__"``: ` `     `  `    ``X, Y ``=` `1``, ``3` ` `  `    ``cost1, cost2 ``=` `391``, ``555` ` `  `    ``makeZero(X, Y, cost1, cost2) ` `    `

## C#

 `// C# implementation to find the minimum ` `// cost to make the two integers equal ` `// to zero using given operations ` `using` `System; ` ` `  `class` `GFG{ ` `     `  `// Function to find out the minimum cost to  ` `// make two number X and Y equal to zero ` `static` `void` `makeZero(``int` `x, ``int` `y, ``int` `a, ``int` `b) ` `{ ` `     `  `    ``// If x is greater than y then swap ` `    ``if``(x > y) ` `    ``{ ` `        ``int` `temp = x; ` `        ``x = y; ` `        ``y = temp; ` `    ``} ` `     `  `    ``// Cost of making y equal to x  ` `    ``int` `tot_cost = (y - x) * a; ` ` `  `    ``// Cost if we choose 1st operation ` `    ``int` `cost1 = 2 * x * a; ` `     `  `    ``// Cost if we choose 2nd operation ` `    ``int` `cost2 = x * b; ` `     `  `    ``// Total cost ` `    ``tot_cost += Math.Min(cost1, cost2); ` `     `  `    ``Console.Write(tot_cost); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int` `X = 1, Y = 3; ` `    ``int` `cost1 = 391, cost2 = 555; ` ` `  `    ``makeZero(X, Y, cost1, cost2); ` `}  ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```1337
```

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

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Improved By : coder001, Code_Mech

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