# 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

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

## C++

`// Simple C++ method to find sum of the` `// elements of array.` `#include <iostream>` `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[0]);` ` ` `cout << arraySum(arr, n);` ` ` `return` `0;` `}` |

## Java

`// 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.` |

## Python3

`# 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#

`// 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.` |

## PHP

`<?php` `// Simple PHP method` `// to find sum of the` `// elements of array.` `function` `arraySum(` `$arr` `, ` `$n` `)` `{` ` ` `$sum` `= 0;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `$sum` `= ` `$sum` `+ ` `$arr` `[` `$i` `];` ` ` `return` `$sum` `;` `}` `// Driver code` `$arr` `= ` `array` `(10, 11, 12, 13,` ` ` `12, 11, 10);` `$n` `= sizeof(` `$arr` `);` `echo` `(arraySum(` `$arr` `, ` `$n` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`<script>` ` ` `// Simple Javascript method` `// to find sum of the` `// elements of array.` `function` `arraySum(arr, n)` `{` ` ` `let sum = 0;` ` ` `for` `(let i = 0; i < n; i++)` ` ` `sum = sum + arr[i];` ` ` `return` `sum;` `}` ` ` `// Driver code` `let arr = [10, 11, 12, 13,` ` ` `12, 11, 10];` `let n = arr.length;` `document.write(arraySum(arr, n));` ` ` `// This code is contributed by _saurabh_jaiswal.` `</script>` |

Output:

79

An **efficient solution **is to apply below formula.

sum = (arr[0] - 1)*n + âŒˆn/2âŒ‰How does it work? If we take a closer look, we can notice that the sum can be written as. (arr[0] - 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[0]-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 = (x^{2}^{2}+ x)/2 + (n^{2}- x)/2 = (2 * x^{2})/2 = x^{2}

## C++

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

## Java

`// 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.` |

## Python3

`# 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#

`// 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[0] - 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.` |

## PHP

`<?php` `// Efficient PHP method to` `// find sum of the elements` `// of array that is halfway` `// increasing and then halfway` `// decreassing` `function` `arraySum(` `$arr` `, ` `$n` `)` `{` ` ` `$x` `= (` `$n` `+ 1) / 2;` ` ` `return` `(` `$arr` `[0] - 1) *` ` ` `$n` `+ ` `$x` `* ` `$x` `;` `}` `// Driver code` `$arr` `= ` `array` `(10, 11, 12, 13,` ` ` `12, 11, 10);` `$n` `= sizeof(` `$arr` `);` `echo` `(arraySum(` `$arr` `, ` `$n` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`// Efficient Javascript method to` `// find sum of the elements` `// of array that is halfway` `// increasing and then halfway` `// decreassing` ` ` `function` `arraySum(arr, n)` `{` ` ` `let x = (n + 1) / 2;` ` ` `return` `(arr[0] - 1) *` ` ` `n + x * x;` `}` ` ` `// Driver code` `let arr = [10, 11, 12, 13,` ` ` `12, 11, 10];` `let n = arr.length;` `document.write(arraySum(arr, n));` ` ` `// This code is contributed by _saurabh_jaiswal.` |

Output:

79

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