Minimize cost required to complete all processes
Given a 2D array arr[][] with each row of the form { X, Y }, where Y and X represents the minimum cost required to initiate a process and the total cost spent to complete the process respectively. The task is to find the minimum cost required to complete all the process of the given array if processes can be completed in any order.
Examples:
Input: arr[][] = { { 1, 2 }, { 2, 4 }, { 4, 8 } }
Output: 8
Explanation:
Consider an initial cost of 8.
Initiate the process arr[2] and after finishing the process, remaining cost = 8 – 4 = 4
Initiate the process arr[1] and after finishing the process, remaining cost = 4 – 2 = 2
Initiate the process arr[0] and after finishing the process, remaining cost = 2 – 1 = 1
Therefore, the required output is 8.Input: arr[][] = { { 1, 7 }, { 2, 8 }, { 3, 9 }, { 4, 10 }, { 5, 11 }, { 6, 12 } }
Output: 27
Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:
- Sort the array in descending order of Y.
- Initialize a variable, say minCost, to store the minimum cost required to complete all the process.
- Initialize a variable, say minCostInit, to store the minimum cost to initiate a process.
- Traverse the array using variable i. For every ith iteration, check if minCostInit is less than arr[i][1] or not. If found to be true then increment the value of minCost by (arr[i][1] – minCostInit) and update minCostInit = arr[i][1].
- In every ith iteration also update the value of minCostInit -= arr[i][0].
- Finally, print the value of minCost.
Below is the implementation of the above approach.
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;bool func(pair<int, int> i1, pair<int, int> i2){ return (i1.first - i1.second < i2.first - i2.second);}// Function to find minimum cost required// to complete all the processint minimumCostReqToCompthePrcess( vector<pair<int, int>> arr){ // Sort the array on descending order of Y sort(arr.begin(), arr.end(), func); // Stores length of array int n = arr.size(); // Stores minimum cost required to // complete all the process int minCost = 0; // Stores minimum cost to initiate // any process int minCostInit = 0; // Traverse the array for(int i = 0; i < n; i++) { // If minCostInit is less than // the cost to initiate the process if (arr[i].second > minCostInit) { // Update minCost minCost += (arr[i].second - minCostInit); // Update minCostInit minCostInit = arr[i].second; } // Update minCostInit minCostInit -= arr[i].first; } // Return minCost return minCost;}// Driver Codeint main(){ vector<pair<int, int>> arr = { { 1, 2 }, { 2, 4 }, { 4, 8 } }; // Function Call cout << (minimumCostReqToCompthePrcess(arr));}// This code is contributed by grand_master |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to find minimum cost required // to complete all the process static int minimumCostReqToCompthePrcess( int[][] arr) { // Sort the array on descending order of Y Arrays.sort(arr, (a, b)->b[1]-a[1]); // Stores length of array int n = arr.length; // Stores minimum cost required to // complete all the process int minCost = 0; // Stores minimum cost to initiate // any process int minCostInit = 0; // Traverse the array for(int i = 0; i < n; i++) { // If minCostInit is less than // the cost to initiate the process if (arr[i][1] > minCostInit) { // Update minCost minCost += (arr[i][1] - minCostInit); // Update minCostInit minCostInit = arr[i][1]; } // Update minCostInit minCostInit -= arr[i][0]; } // Return minCost return minCost; } // Driver code public static void main (String[] args) { int[][] arr = { { 1, 2 }, { 2, 4 }, { 4, 8 } }; // Function Call System.out.println(minimumCostReqToCompthePrcess(arr)); }}// This code is contributed by offbeat |
Python3
# Python3 program to implement# the above approach# Function to find minimum cost required# to complete all the processdef minimumCostReqToCompthePrcess(arr): # Sort the array on descending order of Y arr.sort(key = lambda x: x[0] - x[1]) # Stores length of array n = len(arr) # Stores minimum cost required to # complete all the process minCost = 0 # Stores minimum cost to initiate # any process minCostInit = 0 # Traverse the array for i in range(n): # If minCostInit is less than # the cost to initiate the process if arr[i][1] > minCostInit: # Update minCost minCost += (arr[i][1] - minCostInit) # Update minCostInit minCostInit = arr[i][1] # Update minCostInit minCostInit -= arr[i][0] # Return minCost return minCost# Driver Codeif __name__ == "__main__": arr = [[1, 2], [2, 4], [4, 8]] # Function Call print(minimumCostReqToCompthePrcess(arr)) |
Javascript
<script>// Javascript program to implement// the above approach// Function to find minimum cost required// to complete all the processfunction minimumCostReqToCompthePrcess(arr){ // Sort the array on descending order of Y arr=arr.map(row=>row).reverse() // Stores length of array var n = arr.length; // Stores minimum cost required to // complete all the process var minCost = 0; // Stores minimum cost to initiate // any process var minCostInit = 0; // Traverse the array for(var i = 0; i < n; i++) { // If minCostInit is less than // the cost to initiate the process if (arr[i][1] > minCostInit) { // Update minCost minCost += (arr[i][1] - minCostInit); // Update minCostInit minCostInit = arr[i][1]; } // Update minCostInit minCostInit -= arr[i][0]; } // Return minCost return minCost;}// Driver Codevar arr = [ [ 1, 2 ], [ 2, 4 ], [ 4, 8 ] ];// Function Calldocument.write(minimumCostReqToCompthePrcess(arr));// This code is contributed by rutvik_56.</script> |
8
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
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