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 = 555Output:1337Explanation:

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 = 7Output:4Explanation:

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

**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 <bits/stdc++.h>` `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` |

## Javascript

`<script>` `// Javascript 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` `function` `makeZero(x, y, a, b)` `{` ` ` ` ` `// If x is greater than y then swap` ` ` `if` `(x > y)` ` ` `{` ` ` `let temp = x;` ` ` `x = y;` ` ` `y = temp;` ` ` `}` ` ` `// Cost of making y equal to x` ` ` `let tot_cost = (y - x) * a;` ` ` `// Cost if we choose 1st operation` ` ` `let cost1 = 2 * x * a;` ` ` `// Cost if we choose 2nd operation` ` ` `let cost2 = x * b;` ` ` `// Total cost` ` ` `tot_cost += Math.min(cost1, cost2);` ` ` `document.write(tot_cost);` `}` `// Driver code` `let X = 1, Y = 3;` `let cost1 = 391, cost2 = 555;` `makeZero(X, Y, cost1, cost2);` `// This code is contributed by rameshtravel07 ` `</script>` |

**Output:**

1337

**Time Complexity:** *O(1)* **Auxiliary Space: ***O(1)*

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