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Minimum number of days required to complete the work

  • Last Updated : 03 Jun, 2021

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 of which can be completed. 
Also, it is mentioned that each work has to be completed according to the non-decreasing date of the array D1[].
The task is to find the minimum number of days required to complete the work in non-decreasing order of days in D1[].

Examples

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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 :

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++




// 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;
}

Java




// 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

Javascript




<script>
  
// Javascript program to find the minimum
// number days required
 
// Function to find the minimum
// number days required
function minimumDays(N, D1, D2)
{
    // initialising ans to least value possible
    var ans = -1000000000;
 
    // vector to store the pair of D1(i) and D2(i)
    var vect = [];
 
    for (var i = 0; i < N; i++)
        vect.push([D1[i], D2[i]]);
     
 
    // sort by first i.e D(i)
    vect.sort((a,b)=>a-b)
 
    // Calculate the minimum possible days
    for (var i = 0; i < N; i++) {
        if (vect[i][1] >= ans)
            ans = vect[i][1];
        else
            ans = vect[i][0];
    }
 
    // return the answer
    return ans;
}
 
// Driver Code
// Number of works
var N = 3;
 
// D1[i]
var D1 = [6, 5, 4 ];
 
// D2[i]
var D2 = [1, 2, 3 ];
document.write( minimumDays(N, D1, D2));
 
 
</script>
Output: 
6

 




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