# Minimum time to finish tasks without skipping two consecutive

Given the time taken by n tasks. Find the minimum time needed to finish the tasks such that skipping of tasks is allowed, but can not skip two consecutive tasks.

Examples :

```Input : arr[] = {10, 5, 7, 10}
Output : 12
We can skip first and last task and
finish these task in 12 min.

Input : arr[] = {10}
Output : 0
There is only one task and we can
skip it.

Input : arr[] = {10, 30}
Output : 10

Input : arr[] = {10, 5, 2, 4, 8, 6, 7, 10}
Output : 22```
Recommended Practice

The given problem has the following recursive property.
Let minTime(i) be minimum time to finish till i’th task. It can be written as minimum of two values.

1. Minimum time if i’th task is included in list, let this time be incl(i)
2. Minimum time if i’th task is excluded from result, let this time be excl(i)
`minTime(i) = min(excl(i), incl(i)) `

The result is minTime(n-1) if there are n tasks and indexes start from 0.
incl(i) can be written as below.

```// There are two possibilities
// (a) Previous task is also included
// (b) Previous task is not included
incl(i) = min(incl(i-1), excl(i-1)) +
arr[i] // Since this is inclusive
// arr[i] must be included ```

excl(i) can be written as below.

```// There is only one possibility (Previous task must be
// included as we can't skip consecutive tasks.
excl(i) = incl(i-1)  ```

A simple solution is to make two tables incl[] and excl[] to store times for tasks. Finally, return a minimum of incl[n-1] and excl[n-1]. This solution requires O(n) time and O(n) space.
If we take a closer look, we can notice that we only need incl and excl of the previous job. So we can save space and solve the problem in O(n) time and O(1) space.

Below is the implementation of the above approach:

## C++

 `// C++ program to find minimum time to finish tasks` `// such that no two consecutive tasks are skipped.` `#include ` `using` `namespace` `std;`   `// arr[] represents time taken by n given tasks` `int` `minTime(``int` `arr[], ``int` `n)` `{` `    ``// Corner Cases` `    ``if` `(n <= 0)` `        ``return` `0;`   `    ``// Initialize value for the case when there` `    ``// is only one task in task list.` `    ``int` `incl = arr[0];  ``// First task is included` `    ``int` `excl = 0;       ``// First task is excluded`   `    ``// Process remaining n-1 tasks` `    ``for` `(``int` `i=1; i

## Java

 `// Java program to find minimum time to` `// finish tasks such that no two` `// consecutive tasks are skipped.` `import` `java.io.*;`   `class` `GFG {`   `    ``// arr[] represents time taken by n` `    ``// given tasks` `    ``static` `int` `minTime(``int` `arr[], ``int` `n)` `    ``{` `        ``// Corner Cases` `        ``if` `(n <= ``0``)` `            ``return` `0``;`   `        ``// Initialize value for the case` `        ``// when there is only one task in ` `        ``// task list.` `        ``// First task is included` `        ``int` `incl = arr[``0``];` `        `  `        ``// First task is excluded` `        ``int` `excl = ``0``;     `   `        ``// Process remaining n-1 tasks` `        ``for` `(``int` `i = ``1``; i < n; i++)` `        ``{` `        ``// Time taken if current task is` `        ``// included. There are two ` `        ``// possibilities` `        ``// (a) Previous task is also included` `        ``// (b) Previous task is not included` `        ``int` `incl_new = arr[i] + Math.min(excl,` `                                       ``incl);`   `        ``// Time taken when current task is not` `        ``// included. There is only one ` `        ``// possibility that previous task is ` `        ``// also included.` `        ``int` `excl_new = incl;`   `        ``// Update incl and excl for next` `        ``// iteration` `        ``incl = incl_new;` `        ``excl = excl_new;` `        ``}`   `        ``// Return minimum of two values for ` `        ``// last task` `        ``return` `Math.min(incl, excl);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr1[] = {``10``, ``5``, ``2``, ``7``, ``10``};` `        ``int` `n1 = arr1.length;` `        ``System.out.println(minTime(arr1, n1));`   `        ``int` `arr2[] = {``10``, ``5``, ``7``, ``10``};` `        ``int` `n2 = arr2.length;` `        ``System.out.println(minTime(arr2, n2));`   `        ``int` `arr3[] = {``10``, ``5``, ``2``, ``4``, ``8``, ``6``, ``7``, ``10``};` `        ``int` `n3 = arr3.length;` `        ``System.out.println(minTime(arr3, n3));`   `    ``}` `}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 program to find minimum` `# time to finish tasks such that no` `# two consecutive tasks are skipped.`   `# arr[] represents time` `# taken by n given tasks` `def` `minTime(arr, n):`   `    ``# Corner Cases` `    ``if` `(n <``=` `0``): ``return` `0`   `    ``# Initialize value for the ` `    ``# case when there is only` `    ``# one task in task list.` `    ``incl ``=` `arr[``0``] ``# First task is included` `    ``excl ``=` `0`      `# First task is excluded`   `    ``# Process remaining n-1 tasks` `    ``for` `i ``in` `range``(``1``, n):` `    `  `        ``# Time taken if current task is included` `        ``# There are two possibilities` `        ``# (a) Previous task is also included` `        ``# (b) Previous task is not included` `        ``incl_new ``=` `arr[i] ``+` `min``(excl, incl)`   `        ``# Time taken when current task is not` `        ``# included. There is only one possibility` `        ``# that previous task is also included.` `        ``excl_new ``=` `incl`   `        ``# Update incl and excl for next iteration` `        ``incl ``=` `incl_new` `        ``excl ``=` `excl_new` `    `    `    ``# Return minimum of two values for last task` `    ``return` `min``(incl, excl)`   `# Driver code` `arr1 ``=` `[``10``, ``5``, ``2``, ``7``, ``10``]` `n1 ``=` `len``(arr1)` `print``(minTime(arr1, n1))`   `arr2 ``=` `[``10``, ``5``, ``7``, ``10``]` `n2 ``=` `len``(arr2)` `print``(minTime(arr2, n2))`   `arr3 ``=` `[``10``, ``5``, ``2``, ``4``, ``8``, ``6``, ``7``, ``10``]` `n3 ``=` `len``(arr3)` `print``(minTime(arr3, n3))`   `# This code is contributed by Anant Agarwal.`

## C#

 `// C# program to find minimum time to` `// finish tasks such that no two` `// consecutive tasks are skipped.` `using` `System;`   `class` `GFG {` ` `  `    ``// arr[] represents time taken by n` `    ``// given tasks` `    ``static` `int` `minTime(``int` `[]arr, ``int` `n)` `    ``{` `        ``// Corner Cases` `        ``if` `(n <= 0)` `            ``return` `0;` ` `  `        ``// Initialize value for the case` `        ``// when there is only one task in ` `        ``// task list.` `        ``// First task is included` `        ``int` `incl = arr[0];` `         `  `        ``// First task is excluded` `        ``int` `excl = 0;     ` ` `  `        ``// Process remaining n-1 tasks` `        ``for` `(``int` `i = 1; i < n; i++)` `        ``{` `        ``// Time taken if current task is` `        ``// included. There are two ` `        ``// possibilities` `        ``// (a) Previous task is also included` `        ``// (b) Previous task is not included` `        ``int` `incl_new = arr[i] + Math.Min(excl,` `                                       ``incl);` ` `  `        ``// Time taken when current task is not` `        ``// included. There is only one ` `        ``// possibility that previous task is ` `        ``// also included.` `        ``int` `excl_new = incl;` ` `  `        ``// Update incl and excl for next` `        ``// iteration` `        ``incl = incl_new;` `        ``excl = excl_new;` `        ``}` ` `  `        ``// Return minimum of two values for ` `        ``// last task` `        ``return` `Math.Min(incl, excl);` `    ``}` ` `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[]arr1 = {10, 5, 2, 7, 10};` `        ``int` `n1 = arr1.Length;` `        ``Console.WriteLine(minTime(arr1, n1));` ` `  `        ``int` `[]arr2 = {10, 5, 7, 10};` `        ``int` `n2 = arr2.Length;` `        ``Console.WriteLine(minTime(arr2, n2));` ` `  `        ``int` `[]arr3 = {10, 5, 2, 4, 8, 6, 7, 10};` `        ``int` `n3 = arr3.Length;` `        ``Console.WriteLine(minTime(arr3, n3));` ` `  `    ``}` `}` `// This code is contributed by Anant Agarwal.`

## PHP

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## Javascript

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Output

```12
12
22
```

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

Related Problems:
Find minimum time to finish all jobs with given constraints
Maximum sum such that no two elements are adjacent.
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