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:
- 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 <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 |
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. |
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 |
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)
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