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 i^th 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 i^th 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 process
int 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 Code
int 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 approach
import 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 process
def 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 Code
if __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 process
function 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 Code
var arr = [ [ 1, 2 ],
                               [ 2, 4 ],
                               [ 4, 8 ] ];
// Function Call
document.write(minimumCostReqToCompthePrcess(arr));

// This code is contributed by rutvik_56.
</script>

Output:

8

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