Find minimum sum such that one of every three consecutive elements is taken

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 ` `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]);` `    ``printf``(``"Min Sum is %d"``, findMinSum(arr, n));` `    ``return` `0;` `}`

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 `   `// Find minimum between two numbers.` `int` `min(``int` `num1, ``int` `num2)` `{` `  ``return` `(num1 > num2) ? num2 : num1;` `}`   `// 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]);` `    ``printf``(``"Min Sum is %d"``, findMinSum(arr, n));` `    ``return` `0;` `}`   `// This code is contributed by Aditya Kumar (adityakumar129)`

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

 ``

Javascript

 ``

Output

`Min Sum is 4`

Time Complexity : O(n)
Auxiliary Space : O(n), since n extra space has been taken.

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!