Minimum number of days required to complete the work

Given N works numbered from 1 to N. Given two arrays D1[] and D2[] of N elements each. Also, each work number W(i) is assigned days, D1[i] and D2[i](Such that, D2[i] < D1[i]) either on which it can be completed.

Also, it is mentioned that each work has to completed according to non-decreasing date of the array D1[].

The task is to find the minimum number of days required to complete the work in a non-decreasing order of days in D1[].

Examples:

Input :
N = 3
D1[] = {5, 3, 4}
D2[] = {2, 1, 2}
Output : 2



Explanation:
3 works are to be completed. The first value on Line(i) is D1(i) and the second value is D2(i) where D2(i) < D1(i). The smart worker can finish the second work on Day 1 and then both third work and first work in Day 2, thus maintaining the non-decreasing order of D1[], [3 4 5].

Input :
N = 6
D1[] = {3, 3, 4, 4, 5, 5}
D2[] = {1, 2, 1, 2, 4, 4}
Output : 4

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

Approach: The solution is greedy. The work(i) can be sorted by increasing D1[i], breaking the ties by increasing D2[i]. If we consider the works in this order, we can try to finish the works as early as possible. First of all complete the first work on D2[1]. Move to the second work. If we can complete it on day D2[2] such that (D2[1]<=D2[2]), do it. Otherwise, do the work on day D[2]. Repeat the process until we complete the N-th work, keeping the day of the latest work.

Below is the implementation of the above approach

C++

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// C++ program to find the minimum 
// number days required
  
#include <bits/stdc++.h>
using namespace std;
#define inf INT_MAX
  
// Function to find the minimum 
// number days required
int minimumDays(int N, int D1[], int D2[])
{
    // initialising ans to least value possible
    int ans = -inf;
  
    // vector to store the pair of D1(i) and D2(i)
    vector<pair<int, int> > vect;
  
    for (int i = 0; i < N; i++) 
        vect.push_back(make_pair(D1[i], D2[i]));
      
  
    // sort by first i.e D(i)
    sort(vect.begin(), vect.end());
  
    // Calculate the minimum possible days
    for (int i = 0; i < N; i++) {
        if (vect[i].second >= ans)
            ans = vect[i].second;
        else
            ans = vect[i].first;
    }
  
    // return the answer
    return ans;
}
  
// Driver Code
int main()
{
    // Number of works
    int N = 3;
      
    // D1[i]
    int D1[] = { 6, 5, 4 };
      
    // D2[i]
    int D2[] = { 1, 2, 3 };
  
    cout<<minimumDays(N, D1, D2);
      
    return 0;
}

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Java

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// Java program to find the minimum 
// number days required 
import java.util.*; 
import java.lang.*; 
import java.io.*; 
  
// pair class for number of days
class Pair
{
    int x, y;
      
    Pair(int a, int b)
    {
        this.x = a;
        this.y = b;
    }
}
  
class GFG
{
static int inf = Integer.MIN_VALUE;
  
// Function to find the minimum 
// number days required 
public static int minimumDays(int N, int D1[], 
                              int D2[])
{
    // initialising ans to 
    // least value possible 
    int ans = -inf;
      
    ArrayList<Pair> 
              list = new ArrayList<Pair>();
      
    for (int i = 0; i < N; i++) 
    list.add(new Pair(D1[i], D2[i]));
      
    // sort by first i.e D(i) 
    Collections.sort(list, new Comparator<Pair>() 
    {
        @Override
        public int compare(Pair p1, Pair p2) 
        {
            return p1.x - p2.x;
        }
    });
      
// Calculate the minimum possible days 
for (int i = 0; i < N; i++) 
    if (list.get(i).y >= ans) 
        ans = list.get(i).y; 
    else
        ans = list.get(i).x; 
  
return ans; 
}
  
// Driver Code
public static void main (String[] args) 
{
    // Number of works 
    int N = 3
  
    // D1[i] 
    int D1[] = new int[]{6, 5, 4}; 
      
    // D2[i] 
    int D2[] = new int[]{1, 2, 3};
      
    System.out.print(minimumDays(N, D1, D2));
}
}
  
// This code is contributed by Kirti_Mangal

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Output:

6


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Improved By : Kirti_Mangal