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 high-effort tasks only if you choose 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

Try It Yourself

Method 1:

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.

Below is the implementation of the above approach:

C++

// A naive recursive C++ program to find maximum 
// tasks. 
#include <iostream>
using namespace std;

// 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 code
int main() 
{ 
    int n = 5; 
    int high[] = {3, 6, 8, 7, 6}; 
    int low[] = {1, 5, 4, 5, 3}; 
    cout << maxTasks(high, low, n); 
    return 0; 
} 

// This code is contributed by Shubhamsingh10

// 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("%d\n", maxTasks(high, low, n));
    return 0;
}

Java

// A naive recursive Java program 
// to find maximum tasks.
import java.io.*;
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

// 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.

JavaScript

<script>

// A naive recursive Java 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 Math.max(high[n - 1] + maxTasks(high, low, (n - 2)),
                low[n - 1] + maxTasks(high, low, (n - 1)));
    }


    // Driver code

        let n = 5;
        let high = [3, 6, 8, 7, 6];
        let low = [1, 5, 4, 5, 3];
        document.write( maxTasks(high, low, n));;

</script>

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

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.

Method 2: Dynamic Programming Solution

C++

// A DP based C++ program to find maximum tasks.
#include <iostream>
using namespace std;

// 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};
    cout << maxTasks(high, low, n);
    return 0;
}

// This code is contributed by shivanisinghss2110

// 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("%d", maxTasks(high, low, n));
    return 0;
}

Java

// A DP based Java program to find maximum tasks.
import java.io.*;
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.

Python3

# A DP based Python3 program to find maximum tasks.
# Returns the maximum among the 2 numbers
def max1(x, y):

    return x if(x > y) else y;

# Returns maximum amount of task 
# that can be done till day n
def maxTasks(high, low, n):

    # An array task_dp that stores 
    # the maximum task done
    task_dp = [0] * (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 in range(2, n + 1):
        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 = [3, 6, 8, 7, 6];
low = [1, 5, 4, 5, 3];
print(maxTasks(high, low, n));

# This code is contributed by mits

// 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

JavaScript

<script>

// A DP based javascript program to find maximum tasks.

    
// Returns the maximum among the 2 numbers
function max(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
    var task_dp = Array.from({length: n+1}, (_, i) => 0);

    // 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] = Math.max(high[i - 1] + task_dp[i - 2],
                        low[i - 1] + task_dp[i - 1]);
    return task_dp[n];
}

// Driver code
var n = 5;
var high = [3, 6, 8, 7, 6];
var low = [1, 5, 4, 5, 3];
document.write(maxTasks(high, low, n));

// This code is contributed by Amit Katiyar 
</script>

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

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

Efficient approach: Space optimization

To optimize the space complexity of the above approach, we can eliminate the need for an array task_dp and instead use variables to store the previous maximum values. This will reduce the space complexity from O(n) to O(1).

Implementation Steps:

Define a function maxTasks that takes in two arrays high and low, and the integer n representing the number of days.
Initialize two variables: prev_prev to store the previous-to-previous maximum value, and prev to store the previous maximum value. Set prev to the first element of the high array.
Iterate from i = 2 to n (inclusive) to calculate the maximum amount of tasks that can be done till day n.
Inside the loop, calculate the current maximum value curr using the formula max(high[i - 1] + prev_prev, low[i - 1] + prev).
Update prev_prev with the value of prev, and prev with the value of curr for the next iteration.
After the loop, prev will hold the maximum amount of tasks that can be done till day n.
Return the value of prev.

Implementation:

C++

#include <iostream>
using namespace std;

// 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)
{
    int prev_prev = 0; // Variable to store the previous to previous maximum value
    int prev = high[0]; // Variable to store the previous maximum value

    for (int i = 2; i <= n; i++)
    {
        int curr = max(high[i - 1] + prev_prev, low[i - 1] + prev); // Calculate the current maximum value
        prev_prev = prev; // Update the previous to previous maximum value for the next iteration
        prev = curr; // Update the previous maximum value for the next iteration
    }

    return prev; // Return the maximum value
}

int main()
{
    int n = 5;
    int high[] = {3, 6, 8, 7, 6};
    int low[] = {1, 5, 4, 5, 3};
    cout << maxTasks(high, low, n);
    return 0;
}

Java

import java.util.*;

public class GFG {

    // Returns the maximum among the 2 numbers
    public static int max(int x, int y) {
        return (x > y ? x : y);
    }

    // Returns the maximum amount of tasks that can be done till day n
    public static int maxTasks(int[] high, int[] low, int n) {
        int prev_prev = 0; // Variable to store the previous-to-previous maximum value
        int prev = high[0]; // Variable to store the previous maximum value

        for (int i = 2; i <= n; i++) {
            int curr = max(high[i - 1] + prev_prev, low[i - 1] + prev); // Calculate the current maximum value
            prev_prev = prev; // Update the previous-to-previous maximum value for the next iteration
            prev = curr; // Update the previous maximum value for the next iteration
        }

        return prev; // Return the maximum value
    }

    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));
    }
}

Python3

def max_val(x, y):
    # Return the maximum of two values
    return x if x > y else y

def max_tasks(high, low, n):
    # Initialize variables to keep track of previous two task values
    prev_prev = 0
    prev = high[0]

    # Loop through the tasks starting from the third task (index 2)
    for i in range(2, n + 1):
        # Calculate the maximum value considering two choices: 
        # 1. Doing the current task from the "high" list and skipping the previous task
        # 2. Doing the current task from the "low" list and doing the previous task
        curr = max_val(high[i - 1] + prev_prev, low[i - 1] + prev)
        
        # Update the previous two task values for the next iteration
        prev_prev = prev
        prev = curr

    # Return the maximum value achievable by completing tasks
    return prev

def main():
    n = 5
    high = [3, 6, 8, 7, 6]  # High value for each task
    low = [1, 5, 4, 5, 3]   # Low value for each task
    print(max_tasks(high, low, n))  # Print the maximum achievable value

if __name__ == "__main__":
    main()

using System;

class Program
{
    // Returns the maximum among the two numbers
    static int Max(int x, int y)
    {
        return (x > y ? x : y);
    }

    // Returns the maximum amount of task that can be done till day n
    static int MaxTasks(int[] high, int[] low, int n)
    {
        int prevPrev = 0; // Variable to store the previous to previous maximum value
        int prev = high[0]; // Variable to store the previous maximum value

        for (int i = 2; i <= n; i++)
        {
            int current = Max(high[i - 1] + prevPrev, low[i - 1] + prev); // Calculate the current maximum value
            prevPrev = prev; // Update the previous to previous maximum value for the next iteration
            prev = current; // Update the previous maximum value for the next iteration
        }

        return prev; // Return the maximum value
    }

    static void Main()
    {
        int n = 5;
        int[] high = { 3, 6, 8, 7, 6 };
        int[] low = { 1, 5, 4, 5, 3 };

        // Print the result of maxTasks function
        Console.WriteLine(MaxTasks(high, low, n));
    }
}

JavaScript

// Returns the maximum among the 2 numbers
function max(x, y) {
    return x > y ? x : y;
}

// Returns the maximum amount of tasks that can be done till day n
function maxTasks(high, low, n) {
    let prevPrev = 0; // Variable to store the previous-to-previous maximum value
    let prev = high[0]; // Variable to store the previous maximum value

    for (let i = 2; i <= n; i++) {
        let curr = max(high[i - 1] + prevPrev, low[i - 1] + prev); // Calculate the current maximum value
        prevPrev = prev; // Update the previous-to-previous maximum value for the next iteration
        prev = curr; // Update the previous maximum value for the next iteration
    }

    return prev; // Return the maximum value
}

// Driver code
let n = 5;
let high = [3, 6, 8, 7, 6];
let low = [1, 5, 4, 5, 3];
console.log(maxTasks(high, low, n));

Output

Time Complexity : O(n)
Auxiliary Space: O(1)

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

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