Dynamic Programming | High-effort vs. Low-effort Tasks Problem

You are given n days and for each day (di) you could either perform a high effort tasks (hi) or a low effort tasks (li) or no task with the constraint that you can choose a high-effort tasks only if you chose no task on the previous day. Write a program to find the maximum amount of tasks you can perform within these n days.

Examples:

No. of days (n) = 5
Day      L.E.   H.E
1        1       3
2        5       6
3        4       8
4        5       7
5        3       6
Maximum amount of tasks 
        = 3 + 5 + 4 + 5 + 3 
        = 20

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Optimal Substructure
To find the maximum amount of tasks done till i’th day, we need to compare 2 choices:

Go for high effort tasks on that day, then find the maximum amount of tasks done till (i – 2) th day.
Go for low effort task on that day and find the maximum amount of tasks done till (i – 1) th day.

Let high [1…n] be the input array for high effort task amount on i’th day and low [1…n] be the input array for low effort task amount on ith day.
Let max_task (high [], low [], i) be the function that returns maximum amount of task done till ith day, so it will return max(high[i] + max_task(high, low, (i – 2)), low [i] + max_task (high, low, (i – 1)))

Therefore, the problem has optimal substructure property as the problem can be solved using solutions to subproblems.

Overlapping Subproblems
Following is a simple recursive implementation of the High-effort vs. Low-effort task problem. The implementation simply follows the recursive structure mentioned above. So, High-effort vs. Low-effort Task Problem has both properties of a dynamic programming problem.

C

// A naive recursive C program to find maximum 
// tasks. 
#include<stdio.h> 
  
// Returns the maximum among the 2 numbers 
int max(int x, int y) 
{ 
    return (x > y ? x : y); 
} 
  
// Returns maximum amount of task that can be 
// done till day n 
int maxTasks(int high[], int low[], int n) 
{ 
    // If n is less than equal to 0, then no 
    // solution exists 
    if (n <= 0) 
        return 0; 
  
    /* Determines which task to choose on day n, 
       then returns the maximum till that day */
    return max(high[n-1] + maxTasks(high, low, (n-2)), 
              low[n-1] + maxTasks(high, low, (n-1))); 
} 
  
// Driver program to test above function 
int main() 
{ 
    int n = 5; 
    int high[] = {3, 6, 8, 7, 6}; 
    int low[] = {1, 5, 4, 5, 3}; 
    printf("%dn", maxTasks(high, low, n)); 
    return 0; 
} 

Java

// A naive recursive Java program  
// to find maximum tasks. 
  
class GFG{ 
      
    // Returns maximum amount of task  
    // that can be done till day n 
    static int maxTasks(int high[], int low[], int n) 
    { 
          
        // If n is less than equal to 0, 
        // then no solution exists 
        if (n <= 0) 
            return 0; 
  
        /* Determines which task to choose on day n, 
            then returns the maximum till that day */
        return Math.max(high[n - 1] + maxTasks(high, low, (n - 2)), 
                low[n - 1] + maxTasks(high, low, (n - 1))); 
    } 
  
    // Driver code 
    public static void main(String []args) 
    { 
        int n = 5; 
        int high[] = {3, 6, 8, 7, 6}; 
        int low[] = {1, 5, 4, 5, 3}; 
        System.out.println( maxTasks(high, low, n)); 
    } 
} 
  
// This code is contributed by Ita_c. 

Python3

# A naive recursive Python3 program to  
# find maximum tasks.  
  
# Returns maximum amount of task  
# that can be done till day n  
def maxTasks(high, low, n) : 
      
    # If n is less than equal to 0,  
    # then no solution exists  
    if (n <= 0) : 
        return 0
  
    # Determines which task to choose on day n,  
    # then returns the maximum till that day  
    return max(high[n - 1] + maxTasks(high, low, (n - 2)),  
               low[n - 1] + maxTasks(high, low, (n - 1)));  
  
# Driver Code 
if __name__ == "__main__" :  
  
    n = 5;  
    high = [3, 6, 8, 7, 6]  
    low = [1, 5, 4, 5, 3] 
    print(maxTasks(high, low, n));  
  
# This code is contributed by Ryuga 

C#

// A naive recursive C# program  
// to find maximum tasks. 
using System;  
  
class GFG 
{ 
      
    // Returns maximum amount of task  
    // that can be done till day n 
    static int maxTasks(int[] high, 
                    int[] low, int n) 
    { 
          
        // If n is less than equal to 0, 
        // then no solution exists 
        if (n <= 0) 
            return 0; 
  
        /* Determines which task to choose on day n, 
            then returns the maximum till that day */
        return Math.Max(high[n - 1] +  
            maxTasks(high, low, (n - 2)), low[n - 1] +  
            maxTasks(high, low, (n - 1))); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 5; 
        int[] high = {3, 6, 8, 7, 6}; 
        int[] low = {1, 5, 4, 5, 3}; 
        Console.Write( maxTasks(high, low, n)); 
    } 
} 
  
// This code is contributed by Ita_c. 

Output :

It should be noted that the above function computes the same subproblems again and again.
Therefore, this problem has Overlapping Subproblems Property. So the High-effort vs. Low-effort Task Problem has both the properties of a dynamic programming problem.

Dynamic Programming Solution

// A DP based C++ program to find maximum tasks. 
#include<stdio.h> 
  
// Returns the maximum among the 2 numbers 
int max(int x, int y) 
{ 
    return (x > y ? x : y); 
} 
  
// Returns maximum amount of task that can be 
// done till day n 
int maxTasks(int high[], int low[], int n) 
{ 
    // An array task_dp that stores the maximum 
    // task done 
    int task_dp[n+1]; 
  
    // If n = 0, no solution exists 
    task_dp[0] = 0; 
  
    // If n = 1, high effort task on that day will 
    // be the solution 
    task_dp[1] = high[0]; 
  
    // Fill the entire array determining which 
    // task to choose on day i 
    for (int i = 2; i <= n; i++) 
        task_dp[i] = max(high[i-1] + task_dp[i-2], 
                         low[i-1] + task_dp[i-1]); 
    return task_dp[n]; 
} 
  
// Driver program to test above function 
int main() 
{ 
    int n = 5; 
    int high[] = {3, 6, 8, 7, 6}; 
    int low[] = {1, 5, 4, 5, 3}; 
    printf("%dn", maxTasks(high, low, n)); 
    return 0; 
} 

Output:

Time Complexity : O(n)

This article is contributed by Akash Aggarwal .If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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