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

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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 i^th 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: 7
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 i^th row from the given array or not. The following are the recurrence relation.

MinTime(T_A, T_B, T_C, K, i) = min(MinTime(max(arr[i][0], T_A), max(arr[i][1], T_B), max(arr[i][2], T_C), K-1, i-1), MinTime(T_A, T_B, T_C, K, i-1))

MinTime(T_A, T_B, T_C, 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 T_A+ T_B+ T_C
if i <= 0:
return Infinite

Follow the steps below to solve the problem:

Initialize a variable, say X and Y while traversing each row recursively using the above recurrence relation.
X stores the minimum completion time by selecting the i^th row of the given array.
Y stores the minimum completion time by not selecting the i^th row of the given array.
Find the value of X and Y using the above-mentioned recurrence relation.
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<bits/stdc++.h>
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<<MinTime(arr, N, 0, 0, 0, K);
  return 0;
}

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

<script>
 
// Javascript program to implement  
// the above approach 
 
// Function to get the minimum
// completion time to select
// exactly K items
function Mletime(arr, i, Ta, Tb, Tc, K)
{
      
    // Base case
    if (K == 0)
    {
        return Ta + Tb + Tc;
    }
    if (i == 0)
        return Number.MAX_VALUE;
  
    // Select the ith item
    let X = Mletime(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
    let Y = Mletime(arr, i - 1, Ta,
                    Tb, Tc, K);
  
    return Math.min(X, Y);
}
 
// Driver code
K = 2;
let n = 3;
let arr = [ [ 1, 2, 2 ],
            [ 3, 4, 1 ],
            [ 3, 1, 2 ] ];
 
document.write(Mletime(arr, n, 0, 0, 0, K));
 
// This code is contributed by splevel62  
 
</script>

Output

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

Approach 2 :-

C++

#include <iostream>
#include <bits/stdc++.h>
 
using namespace std;
int MinTime(vector<vector<int>>& tasks, int k) 
{
    // if k == 0 then return 0
    if (k == 0) 
    {
        return 0;
    }
    vector<pair<int, pair<int, int>>> 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<pair<int, int>> 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<int> 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<vector<int>> arr = 
                  {{1, 2, 2} ,
                   {3, 4, 1} ,
                   {3, 1, 2}
                  };
    
  cout<<MinTime(arr,K)<<endl;
  //This code is given by Akshay Verma
  return 0;
}

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<int[]> 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<int[]> 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<Integer> 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<List<int>> arr = new List<List<int>>();
  
    // make a pair of pair DS
    for(int i = 0; i < tasks.GetLength(0); i++)
    {
        arr.Add(new List<int>());
        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<List<int>> pr = new List<List<int>>();
  
      // push the pairs into the pr vector
      for (int j = 0; j <= i; j++)
      {
        pr.Add(new List<int>());
        pr[j].Add(arr[j][1]);
        pr[j].Add(arr[j][2]);
      }
  
      // priority queue
      List<int> q = new List<int>();
      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

<script>
// Javascript program to implement 
// the above approach
 
function MinTime(tasks,k)
{
    // if k == 0 then return 0
    if (k == 0)
    {
      return 0;
    }
    let arr = [];
  
    // make a pair of pair DS
    for (let t=0;t< tasks.length;t++)
    {
      arr.push([tasks[t][0], tasks[t][1], tasks[t][2]]);
    }
  
    // sort the pairs
    arr.sort(function(a,b){return a[0]-b[0];});
  
    // initialize the ans as INT_MAX/2
    let ans =Number.MAX_VALUE / 2;
    for (let i = k - 1; i < arr.length; i++)
    {
      let pr = [];
  
      // push the pairs into the pr vector
      for (let j = 0; j <= i; j++)
      {
        pr.push([arr[j][1],arr[j][2]]);
      }
  
      // sort the pairs
      pr.sort(function(a,b){return a[0]-b[0];});
  
      // priority queue
      let q = [];
      for (let j = 0; j < pr.length; j++)
      {
        q.push(pr[j][1]);
        if (q.length > k)
        {
          q.shift();
        }
        if (q.length == k)
        {
          ans = Math.min(ans, arr[i][0] +
                         q[0] + pr[j][0]);
        }
      }
    }
    return ans;
}
 
// Driver code
let K = 2;
let n = 3;
let arr=[[ 1, 2, 2 ],
                   [ 3, 4, 1 ],
                   [ 3, 1, 2 ]];
document.write(MinTime(arr,K));
 
 
// This code is contributed by patel2127
</script>

Output

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

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