Given an array of **n** non-negative numbers, the task is to find the minimum sum of elements (picked from the array) such that at least one element is picked out of every 3 consecutive elements in the array.

Examples:

Input : arr[] = {1, 2, 3} Output : 1 Input : arr[] = {1, 2, 3, 6, 7, 1} Output : 4 We pick 3 and 1 (3 + 1 = 4) Note that there are following subarrays of three consecutive elements {1, 2, 3}, {2, 3, 6}, {3, 6, 7} and {6, 7, 1} We have picked one element from every subarray. Input : arr[] = {1, 2, 3, 6, 7, 1, 8, 6, 2, 7, 7, 1} Output : 7 The result is obtained as sum of 3 + 1 + 2 + 1

Let sum(i) be the minimum possible sum when arr[i] is part of a solution sum (not necessarily result) and is last picked element. Then our result is minimum of sum(n-1), sum(n-2) and sum(n-3) [We must pick at least one of the last three elements].

We can recursively compute sum(i) as sum of arr[i] and minimum(sum(i-1), sum(i-2), sum(i-3)). Since there are overlapping subproblems in recursive structure of problem, we can use Dynamic Programming to solve this problem.

Below is C++ implementation of above idea.

// A Dynamic Programming based C++ program to // find minimum possible sum of elements of array // such that an element out of every three // consecutive is picked. #include <iostream> using namespace std; // A utility function to find minimum of // 3 elements int minimum(int a, int b, int c) { return min(min(a, b), c); } // Returns minimum possible sum of elements such // that an element out of every three consecutive // elements is picked. int findMinSum(int arr[], int n) { // Create a DP table to store results of // subpriblems. sum[i] is going to store // minimum possible sum when arr[i] is // part of the solution. int sum[n]; // When there are less than or equal to // 3 elements sum[0] = arr[0]; sum[1] = arr[1]; sum[2] = arr[2]; // Iterate through all other elements for (int i=3; i<n; i++) sum[i] = arr[i] + minimum(sum[i-3], sum[i-2], sum[i-1]); return minimum(sum[n-1], sum[n-2], sum[n-3]); } // Driver code int main() { int arr[] = {1, 2, 3, 20, 2, 10, 1}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Min Sum is " << findMinSum(arr, n); return 0; }

Output:

Min Sum is 4

Time Complexity : O(n)

Auxiliary Space : O(n)

This problem and solution are contributed by **Ayush Saluja**. 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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