Given 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

Expected Time Complexity is O(n) and extra space is O(1).

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.

- Minimum time if i’th task is included in list, let this time be incl(i)
- Minimum time if i’th task is excluded from result, let this time be excl(i)

minTime(i) = min(excl(i), incl(i))

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 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 previous job. So we can save space and solve the problem in O(n) time and O(1) space. Below is C++ implementation of the idea.

## C++

// C++ program to find minimum time to finish tasks // such that no two consecutive tasks are skipped. #include <bits/stdc++.h> 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 exluded // 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] + 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 maximum of two values for last task return min(incl, excl); } // Driver code int main() { int arr1[] = {10, 5, 2, 7, 10}; int n1 = sizeof(arr1)/sizeof(arr1[0]); cout << minTime(arr1, n1) << endl; int arr2[] = {10, 5, 7, 10}; int n2 = sizeof(arr2)/sizeof(arr2[0]); cout << minTime(arr2, n2) << endl; int arr3[] = {10, 5, 2, 4, 8, 6, 7, 10}; int n3 = sizeof(arr3)/sizeof(arr3[0]); cout << minTime(arr3, n3) << endl; return 0; }

## 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 exluded 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 maximum 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

Output :

12 12 22

**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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