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 the implementation of above idea.

## C++

// 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 // subproblems. 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; }

## Java

// A Dynamic Programming based java program to // find minimum possible sum of elements of array // such that an element out of every three // consecutive is picked. import java.io.*; class GFG { // A utility function to find minimum of // 3 elements static int minimum(int a, int b, int c) { return Math. min(Math.min(a, b), c); } // Returns minimum possible sum of elements such // that an element out of every three consecutive // elements is picked. static int findMinSum(int arr[], int n) { // Create a DP table to store results of // subproblems. sum[i] is going to store // minimum possible sum when arr[i] is // part of the solution. int sum[] =new int[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 public static void main (String[] args) { int arr[] = {1, 2, 3, 20, 2, 10, 1}; int n = arr.length; System.out.println("Min Sum is " + findMinSum(arr, n)); } } // This code is contributed by vt_m

## Python3

# A Dynamic Programming based python 3 program to # find minimum possible sum of elements of array # such that an element out of every three # consecutive is picked. # A utility function to find minimum of # 3 elements def minimum(a, b, 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. def findMinSum(arr,n): # Create a DP table to store results of # subproblems. sum[i] is going to store # minimum possible sum when arr[i] is # part of the solution. sum = [] # When there are less than or equal to # 3 elements sum.append(arr[0]) sum.append(arr[1]) sum.append(arr[2]) # Iterate through all other elements for i in range(3, n): sum.append( 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 arr = [1, 2, 3, 20, 2, 10, 1] n = len(arr) print( "Min Sum is ",findMinSum(arr, n)) # This code is contributed by Sam007

## C#

// 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. using System; class GFG { // A utility function to find minimum of // 3 elements static int minimum(int a, int b, int c) { return Math. Min(Math.Min(a, b), c); } // Returns minimum possible sum of elements such // that an element out of every three consecutive // elements is picked. static int findMinSum(int []arr, int n) { // Create a DP table to store results of // subproblems. sum[i] is going to store // minimum possible sum when arr[i] is // part of the solution. int []sum =new int[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 public static void Main () { int []arr = {1, 2, 3, 20, 2, 10, 1}; int n = arr.Length; Console.WriteLine("Min Sum is " + findMinSum(arr, n)); } } //This code is contributed by Sam007

## PHP

<?php // A Dynamic Programming based // PHP program to find minimum // possible sum of elements of // array such that an element // out of every three consecutive // is picked. // function to find minimum of // 3 elements function minimum($a, $b, $c) { return min(min($a, $b), $c); } // Returns minimum possible sum // of elements such that an // element out of every three // consecutive elements is picked. function findMinSum($arr, $n) { // Create a DP table to store results of // subproblems. sum[i] is going to store // minimum possible sum when arr[i] is // part of the solution. $sum[$n] = 0; // When there are less // than or equal to // 3 elements $sum[0] = $arr[0]; $sum[1] = $arr[1]; $sum[2] = $arr[2]; // Iterate through all other elements for ($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 $arr = array(1, 2, 3, 20, 2, 10, 1); $n = sizeof($arr); echo "Min Sum is " , findMinSum($arr, $n); // This code is contributed by nitin mittal. ?>

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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