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

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. Move to the second work. If we can complete it on day D2 such that (D2<=D2), do it. Otherwise, do the work on day D. 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 using namespace std;#define inf INT_MAX // Function to find the minimum// number days requiredint 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 > 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 Codeint main(){    // Number of works    int N = 3;         // D1[i]    int D1[] = { 6, 5, 4 };         // D2[i]    int D2[] = { 1, 2, 3 };     cout<

Java

 // Java program to find the minimum// number days requiredimport java.util.*;import java.lang.*;import java.io.*; // pair class for number of daysclass 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 requiredpublic static int minimumDays(int N, int D1[],                              int D2[]){    // initialising ans to    // least value possible    int ans = -inf;         ArrayList              list = new ArrayList();         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()    {        @Override        public int compare(Pair p1, Pair p2)        {            return p1.x - p2.x;        }    });     // Calculate the minimum possible daysfor (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 Codepublic 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


Output:
6

My Personal Notes arrow_drop_up