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# 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 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 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 > 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.io.*;``import` `java.lang.*;``import` `java.util.*;` `// 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 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 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

 ``

## Python3

 `def` `minDaysWork(n, day1, day2):``    ``ans ``=` `0` `    ``# List to store the pair of day1(i) and day2(i)``    ``combined ``=` `[[] ``for` `i ``in` `range``(n)]` `    ``for` `i ``in` `range``(``len``(day1)):``        ``combined[i] ``=` `[day1[i], day2[i]]` `    ``# Sort the array``    ``combined.sort()``    ``for` `i ``in` `range``(``len``(day1)):``        ``if` `(combined[i][``1``] >``=` `ans):``            ``ans ``=` `combined[i][``1``]``        ``else``:``            ``ans ``=` `combined[i][``0``]` `    ``return` `ans`  `# Driver Code``# Input taken``N ``=` `3``D1 ``=` `[``6``, ``5``, ``4``]``D2 ``=` `[``1``, ``2``, ``3``]``print``(minDaysWork(N, D1, D2))`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `// pair class for number of days``class` `Pair {``    ``public` `int` `x, y;` `    ``public` `Pair(``int` `a, ``int` `b) {``        ``this``.x = a;``        ``this``.y = b;``    ``}``}` `class` `GFG {``    ``static` `int` `inf = ``int``.MinValue;` `    ``// 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;` `        ``List list = ``new` `List();` `        ``for` `(``int` `i = 0; i < N; i++)``            ``list.Add(``new` `Pair(D1[i], D2[i]));` `        ``// sort by first i.e D(i)``        ``list.Sort((p1, p2) => p1.x - p2.x);` `        ``// Calculate the minimum possible days``        ``for` `(``int` `i = 0; i < N; i++) {``            ``if` `(list[i].y >= ans)``                ``ans = list[i].y;``            ``else``                ``ans = list[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 };` `        ``Console.Write(MinimumDays(N, D1, D2));``    ``}``}`

Output:

`6`

Time Complexity: O(nlogn)
Auxiliary Space: O(n)

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