Minimum time required to complete exactly K tasks based on given order of task execution

Last Updated : 09 Mar, 2022

Given a 2D array, arr[][] of size N, each row of which consists of the time required to complete 3 different types of tasks A, B, and C. The row elements arr[i][0], arr[i][1] and arr[i][2] are the time required to complete the ith task of type A, B and C respectively. The task is to find the minimum completion time to complete exactly K tasks from the array based on the following conditions:

• The same type of task will operate at the same time.
• Tasks of type B can be performed only when K tasks of type A have already been completed.
• Tasks of type C can be performed only when K tasks of type B have already been completed.

Examples:

Input: N = 3, K = 2, arr[][] = {{1, 2, 2}, {3, 4, 1}, {3, 1, 2}}
Output:
Explanation:
Minimum time required to complete K tasks of type A = min(max(arr[0][0], arr[2][0]), max(arr[0][0], arr[1][0]), max(arr[1][0], arr[2][0])) = min(max(1, 3), max(1, 3), max(3, 3)) 3
Minimum time required to complete K tasks of type B = max(arr[0][1], arr[2][1]) = 2
Therefore, select items 1 and 3 as the K tasks to be completed for minimum time.
Therefore, time to complete K tasks of type C = max(arr[0][2], arr[2][2]) = 2
Therefore, minimum time to complete K(= 2) of all types = 3 + 2 + 2 = 7

Input: N = 3, K = 3, arr[][] = {{2, 4, 5}, {4, 5, 4}, {3, 4, 5}}
Output: 14

Approach: The problem can be solved using Recursion. The idea is to either select the ith row from the given array or not. The following are the recurrence relation.

MinTime(TA, TB, TC, K, i) = min(MinTime(max(arr[i][0], TA), max(arr[i][1], TB), max(arr[i][2], TC), K-1, i-1), MinTime(TA, TB, TC, K, i-1))

MinTime(TA, TB, TC, K, i) Stores the minimum time to complete K tasks.
i stores the position of the ith row of the given array elements.

Base Case: if K == 0:
return TA + TB + TC
if i <= 0:
return Infinite

Follow the steps below to solve the problem:

1. Initialize a variable, say X and Y while traversing each row recursively using the above recurrence relation.
2. X stores the minimum completion time by selecting the ith row of the given array.
3. Y stores the minimum completion time by not selecting the ith row of the given array.
4. Find the value of X and Y using the above-mentioned recurrence relation.
5. Finally, return the value of min(X, Y) for every row.

Below is the implementation of the above approach:

C++

 // C++ program to implement // the above approach #include using namespace std; #define N 3   // Function to get the minimum // completion time to select // exactly K items int MinTime(int arr[][N],int i,                  int Ta, int Tb,                 int Tc, int K) {     // Base case     if(K == 0) {         return Ta + Tb + Tc;     }     if(i == 0)         return INT_MAX;               // Select the ith item     int X = MinTime(arr, i - 1,             max(Ta, arr[i - 1][0]),             max(Tb, arr[i - 1][1]),             max(Tc, arr[i - 1][2]), K -1);           // Do not select the ith item     int Y = MinTime(arr, i - 1,                     Ta, Tb, Tc, K);                           return min(X, Y); }   // Driver Code int main() {   int K = 2;   int n = 3;   int arr[N][N] = {{1, 2, 2} ,                    {3, 4, 1} ,                    {3, 1, 2}                   };       cout<

Java

 // Java program to implement  // the above approach import java.util.*;   class GFG{   // Function to get the minimum // completion time to select // exactly K items public static int MinTime(int arr[][], int i,                           int Ta, int Tb,                           int Tc, int K) {           // Base case     if (K == 0)     {         return Ta + Tb + Tc;     }     if (i == 0)         return Integer.MAX_VALUE;       // Select the ith item     int X = MinTime(arr, i - 1,                     Math.max(Ta, arr[i - 1][0]),                     Math.max(Tb, arr[i - 1][1]),                     Math.max(Tc, arr[i - 1][2]), K - 1);       // Do not select the ith item     int Y = MinTime(arr, i - 1, Ta,                     Tb, Tc, K);       return Math.min(X, Y); }   // Driver Code public static void main(String args[]) {     int K = 2;     int n = 3;     int arr[][] = { { 1, 2, 2 },                     { 3, 4, 1 },                     { 3, 1, 2 } };       System.out.println(MinTime(arr, n, 0, 0, 0, K)); } }   // This code is contributed by hemanth gadarla

Python3

 # Python3 program to implement # the above approach N = 3   # Function to get the minimum # completion time to select # exactly K items def MinTime(arr, i, Ta, Tb, Tc, K):           # Base case     if (K == 0):         return Ta + Tb + Tc     if (i == 0):         return 10**9       # Select the ith item     X = MinTime(arr, i - 1,             max(Ta, arr[i - 1][0]),             max(Tb, arr[i - 1][1]),             max(Tc, arr[i - 1][2]), K -1)       # Do not select the ith item     Y = MinTime(arr, i - 1,                 Ta, Tb, Tc, K)       return min(X, Y)   # Driver Code if __name__ == '__main__':           K = 2     n = 3     arr = [ [ 1, 2, 2 ],             [ 3, 4, 1 ],             [ 3, 1, 2 ] ]       print(MinTime(arr, N, 0, 0, 0, K))   # This code is contributed by mohit kumar 29

C#

 // C# program to implement  // the above approach using System; class GFG{   // Function to get the minimum // completion time to select // exactly K items public static int MinTime(int [,]arr, int i,                           int Ta, int Tb,                           int Tc, int K) {      // Base case   if (K == 0)   {     return Ta + Tb + Tc;   }   if (i == 0)     return int.MaxValue;     // Select the ith item   int X = MinTime(arr, i - 1,                   Math.Max(Ta,                            arr[i - 1, 0]),                   Math.Max(Tb,                            arr[i - 1, 1]),                   Math.Max(Tc,                            arr[i - 1, 2]),                                K - 1);     // Do not select the ith item   int Y = MinTime(arr, i - 1, Ta,                   Tb, Tc, K);     return Math.Min(X, Y); }   // Driver Code public static void Main(String []args) {   int K = 2;   int n = 3;   int [,]arr = {{1, 2, 2},                 {3, 4, 1},                 {3, 1, 2}};     Console.WriteLine(MinTime(arr, n, 0,                             0, 0, K)); } }   // This code is contributed by shikhasingrajput

Javascript



Output

7

Time Complexity: O(2N)
Auxiliary Space: O(N)

Approach 2 :-

C++

 #include #include   using namespace std; int MinTime(vector>& tasks, int k) {     // if k == 0 then return 0     if (k == 0)     {         return 0;     }     vector>> arr;         // make a pair of pair DS     for (auto t : tasks)     {         arr.push_back({t[0], {t[1], t[2]}});     }         // sort the pairs     sort(arr.begin(), arr.end());         // initialize the ans as INT_MAX/2     int ans = INT_MAX / 2;     for (int i = k - 1; i < arr.size(); i++)     {         vector> pr;                 // push the pairs into the pr vector         for (int j = 0; j <= i; j++)         {             pr.push_back(arr[j].second);         }                 // sort the pairs         sort(pr.begin(), pr.end());                  // priority queue         priority_queue q;         for (int j = 0; j < pr.size(); j++)         {             q.push(pr[j].second);             if (q.size() > k)             {                 q.pop();             }             if (q.size() == k)             {                 ans = min(ans, arr[i].first + q.top() + pr[j].first);             }         }     }     return ans; }   // Driver Code int main() {      int K = 2;   int n = 3;   vector> arr =                   {{1, 2, 2} ,                    {3, 4, 1} ,                    {3, 1, 2}                   };        cout<

Java

 // Java program to implement  // the above approach import java.util.*; class GFG {   static int MinTime(int[][] tasks, int k)   {       // if k == 0 then return 0     if (k == 0)     {       return 0;     }     ArrayList arr = new ArrayList<>();       // make a pair of pair DS     for (int[] t : tasks)     {       arr.add(new int[]{t[0], t[1], t[2]});     }       // sort the pairs     Collections.sort(arr, (a, b)->a[0] - b[0]);       // initialize the ans as INT_MAX/2     int ans =Integer.MAX_VALUE / 2;     for (int i = k - 1; i < arr.size(); i++)     {       ArrayList pr = new ArrayList<>();         // push the pairs into the pr vector       for (int j = 0; j <= i; j++)       {         pr.add(new int[]{arr.get(j)[1],arr.get(j)[2]});       }         // sort the pairs       Collections.sort(pr, (a, b)->a[0] - b[0]);         // priority queue       PriorityQueue q = new PriorityQueue<>();       for (int j = 0; j < pr.size(); j++)       {         q.add(pr.get(j)[1]);         if (q.size() > k)         {           q.poll();         }         if (q.size() == k)         {           ans = Math.min(ans, arr.get(i)[0] +                          q.peek() + pr.get(j)[0]);         }       }     }     return ans;   }     // Driver code   public static void main (String[] args)   {     int K = 2;     int n = 3;     int arr[][] = { { 1, 2, 2 },                    { 3, 4, 1 },                    { 3, 1, 2 } };       System.out.println(MinTime(arr,K));     } }   // This code is contributed by offbeat

Python3

 # Python program to implement  # the above approach import sys   def MinTime(tasks,k):        # if k == 0 then return 0     if (k == 0):         return 0           arr = []            # make a pair of pair DS     for t in tasks:         arr.append([t[0], t[1], t[2]])            # sort the pairs     arr.sort()           # initialize the ans as INT_MAX/2     ans = sys.maxsize//2           for i in range(k - 1, len(arr)):         pr = []           # push the pairs into the pr vector         for j in range(i+1):             pr.append([arr[j][1],arr[j][2]])                    # sort the pairs         pr.sort()                   # priority queue         q = []                   for j in range(len(pr)):             q.append(pr[j][1])                           if(len(q) > k):                 q.pop(0)             if(len(q) == k):                 ans=min(ans, arr[i][0] + q[0] + pr[j][0])     return ans     # Driver code K = 2 n = 3   arr = [[1, 2, 2], [ 3, 4, 1 ], [3, 1, 2 ]] print(MinTime(arr, K))   # This code is contributed by unknown2108

C#

 // C# program to implement // the above approach using System; using System.Collections.Generic; class GFG {        static int MinTime(int[,] tasks, int k)   {        // if k == 0 then return 0     if (k == 0)     {       return 0;     }     List> arr = new List>();        // make a pair of pair DS     for(int i = 0; i < tasks.GetLength(0); i++)     {         arr.Add(new List());         for(int j = 0; j < tasks.GetLength(1); j++)         {          arr[i].Add(tasks[i,j]);           }     }        // initialize the ans as INT_MAX/2     int ans = Int32.MaxValue / 2;     for (int i = k - 1; i < arr.Count; i++)     {       List> pr = new List>();          // push the pairs into the pr vector       for (int j = 0; j <= i; j++)       {         pr.Add(new List());         pr[j].Add(arr[j][1]);         pr[j].Add(arr[j][2]);       }          // priority queue       List q = new List();       for (int j = 0; j < pr.Count; j++)       {         q.Add(pr[j][1]);         q.Sort();         q.Reverse();         if (q.Count > k)         {           q.RemoveAt(0);         }         if (q.Count == k)         {           ans = Math.Min(ans, arr[i][0] + q[0] + pr[j][0]) + 1;         }       }     }     return ans;   }     // Driver code   static void Main() {     int K = 2;     int[,] arr = { { 1, 2, 2 },                    { 3, 4, 1 },                    { 3, 1, 2 } };        Console.Write(MinTime(arr,K));   } }   // This code is contributed by decode2207.

Javascript



Output

7

Time Complexity :- O(N2logN)
Space Complexity:- O(N)

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