# Sum of array elements that is first continuously increasing then decreasing

Given an array where elements are first continuously increasing and after that its continuously decreasing unit first number is reached again. We want to add the elements of array. We may assume that there is no overflow in sum.

Examples:

```Input  : arr[] = {5, 6, 7, 6, 5}.
Output : 29

Input  : arr[] = {10, 11, 12, 13, 12, 11, 10}
Output : 79
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to traverse through n and add the elements of array.

 `// Simple C++ method to find sum of the ` `// elements of array. ` `#include ` `using` `namespace` `std; ` `int` `arraySum(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `sum = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``sum = sum + arr[i]; ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {10, 11, 12, 13, 12, 11, 10}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << arraySum(arr, n); ` `    ``return` `0; ` `} `

 `// JAVA Code for Sum of array elements ` `// that is first continuously increasing  ` `// then decreasing ` `class` `GFG { ` `     `  `    ``public` `static` `int` `arraySum(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `sum = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``sum = sum + arr[i]; ` `        ``return` `sum; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``10``, ``11``, ``12``, ``13``, ``12``, ``11``, ``10``}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(arraySum(arr, n)); ` `             `  `    ``} ` `} ` `// This code is contributed by Arnav Kr. Mandal. `

 `# Simple python method to find sum of the ` `# elements of array. ` `def` `arraySum( arr, n): ` `    ``_sum ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``_sum ``=` `_sum ``+` `arr[i] ` `    ``return` `_sum ` ` `  `# Driver code ` `arr ``=` `[``10``, ``11``, ``12``, ``13``, ``12``, ``11``, ``10``] ` `n ``=` `len``(arr) ` `print``(arraySum(arr, n)) ` ` `  `# This code is contributedc by "Abhishek Sharma 44" `

 `// C# Code for Sum of array elements ` `// that is first continuously increasing  ` `// then decreasing ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``public` `static` `int` `arraySum(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `sum = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``sum = sum + arr[i]; ` `        ``return` `sum; ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {10, 11, 12, 13, 12, 11, 10}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(arraySum(arr, n)); ` `             `  `    ``} ` `} ` `// This code is contributed by vt_m. `

 ` `

Output:

```79
```

An efficient solution is to apply below formula.

```sum = (arr - 1)*n + ⌈n/2⌉2

How does it work?
If we take a closer look, we can notice that the
sum can be written as.

(arr - 1)*n + (1 + 2 + .. x + (x -1) + (x-2) + ..1)
Let us understand above result with example {10, 11,
12, 13, 12, 11, 10}.  If we subtract 9 (arr-1) from
this array, we get {1, 2, 3, 2, 1}.

Where x = ceil(n/2)  [Half of array size]

As we know that 1 + 2 + 3 + . . . + x = x * (x + 1)/2.
And we have given
= 1 + 2 + 3 + . . . + x + (x - 1) + . . . + 3 + 2 + 1
= (1 + 2 + 3 + . . . + x) + ((x - 1) + . . . + 3 + 2 + 1)
= (x * (x + 1))/2 + ((x - 1) * x)/2
= (x2 + x)/2 + (n2 - x)/2
= (2 * x2)/2
= x2```
 `// Efficient C++ method to find sum of the ` `// elements of array that is halfway increasing ` `// and then halfway decreassing ` `#include ` `using` `namespace` `std; ` ` `  `int` `arraySum(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `x = (n+1)/2; ` `    ``return` `(arr - 1)*n + x*x; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {10, 11, 12, 13, 12, 11, 10}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << arraySum(arr, n); ` `    ``return` `0; ` `} `

 `// JAVA Code for Sum of array elements ` `// that is first continuously increasing  ` `// then decreasing ` `class` `GFG { ` `     `  `    ``public` `static` `int` `arraySum(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `x = (n + ``1``) / ``2``; ` `        ``return` `(arr[``0``] - ``1``) * n + x * x; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``10``, ``11``, ``12``, ``13``, ``12``, ``11``, ``10``}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(arraySum(arr, n));    ` `    ``} ` `} ` `// This code is contributed by Arnav Kr. Mandal. `

 `# Efficient python method to find sum of the ` `# elements of array that is halfway increasing ` `# and then halfway decreassing ` `def` `arraySum( arr, n): ` `    ``x ``=` `(n ``+` `1``)``/``2` `    ``return` `(arr[``0``] ``-` `1``)``*``n ``+` `x ``*` `x ` `     `  `# Driver code ` `arr ``=` `[``10``, ``11``, ``12``, ``13``, ``12``, ``11``, ``10``] ` `n ``=` `len``(arr) ` `print``(arraySum(arr, n)) ` ` `  `# This code is contributedc by "Abhishek Sharma 44" `

 `// C# Code for Sum of array elements ` `// that is first continuously increasing  ` `// then decreasing ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``public` `static` `int` `arraySum(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `x = (n + 1) / 2; ` `        ``return` `(arr - 1) * n + x * x; ` `    ``} ` `     `  `    ``/* Driver program to test above function */` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {10, 11, 12, 13, 12, 11, 10}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(arraySum(arr, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

 ` `

Output:
```79
```

This article is contributed by Dharmendra kumar. 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.