# 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<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; ` `} ` |

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## 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. ` |

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## 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 ` |

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## 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. ` |

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**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 **

`// 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; ` `} ` |

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Output:

20

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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