# 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 <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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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## 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. ` `?> ` |

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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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