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
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> |
1337
Time Complexity: O(1)
Auxiliary Space: O(1)