Tutorialspoint.dev

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




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

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

PHP

<?php
// A naive recursive PHP program to find maximum
// tasks.
  
// Returns maximum amount of task that can be
// done till day n
function 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
$n = 5;
$high = array(3, 6, 8, 7, 6);
$low = array(1, 5, 4, 5, 3);
print(maxTasks($high, $low, $n));
      
// This code is contributed by mits
?>


Output :

20

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

C++

// 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;
}

Java

// A DP based Java program to find maximum tasks.
class GFG
{
      
// Returns the maximum among the 2 numbers
static int max(int x, int y)
{
    return (x > y ? x : y);
}
  
// Returns maximum amount of task that can be
// done till day n
static int maxTasks(int []high, int []low, int n)
{
    // An array task_dp that stores the maximum
    // task done
    int[] task_dp = new int[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] = Math.max(high[i - 1] + task_dp[i - 2],
                        low[i - 1] + task_dp[i - 1]);
    return task_dp[n];
}
  
// 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 Code_Mech.

C#

// A DP based C# program to find maximum tasks.
using System;
  
class GFG
{
      
// Returns the maximum among the 2 numbers
static int max(int x, int y)
{
    return (x > y ? x : y);
}
  
// Returns maximum amount of task that can be
// done till day n
static int maxTasks(int []high, int []low, int n)
{
    // An array task_dp that stores the maximum
    // task done
    int[] task_dp = new int[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
static void Main()
{
    int n = 5;
    int []high = {3, 6, 8, 7, 6};
    int []low = {1, 5, 4, 5, 3};
    Console.WriteLine(maxTasks(high, low, n));
}
}
  
// This code is contributed by mits

PHP

<?php
// A DP based PHP program to find maximum tasks.
// Returns the maximum among the 2 numbers
function max1($x, $y)
{
    return ($x > $y ? $x : $y);
}
  
// Returns maximum amount of task that can be
// done till day n
function maxTasks($high, $low, $n)
{
    // An array task_dp that stores the maximum
    // task done
    $task_dp = array($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 ($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 code
{
    $n = 5;
    $high = array(3, 6, 8, 7, 6);
    $low = array(1, 5, 4, 5, 3);
    echo(maxTasks($high, $low, $n));
}
  
// This code is contributed by Code_Mech.


Output:

20

Time Complexity : O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter