# Program to find smallest difference of angles of two parts of a given circle

• Last Updated : 22 Jun, 2022

Given a division of a circle into n pieces as an array of size n. The ith element of the array denotes the angle of one piece. Our task is to make two continuous parts from these pieces so that the difference between angles of these two parts is minimum.
Examples :

```Input : arr[] = {90, 90, 90, 90}
Output : 0
In this example, we can take 1 and 2
pieces and 3 and 4 pieces. Then the
answer is |(90 + 90) - (90 + 90)| = 0.

Input : arr[] = {170, 30, 150, 10}
Output : 0
In this example, we can take 1 and 4,
and 2 and 3 pieces. So the answer is
|(170 + 10) - (30 + 150)| = 0.

Input : arr[] = {100, 100, 160}
Output : 40```

We can notice that if one of the parts is continuous then all the remaining pieces also form a continuous part. If angle of the first part is equal to x then difference between angles of first and second parts is |x – (360 – x)| = |2 * x – 360| = 2 * |x – 180|. So for each possible continuous part, we can count its angle and update the answer.

## C++

 `// CPP program to find minimum``// difference of angles of two``// parts of given circle.``#include ``using` `namespace` `std;` `// Returns the minimum difference``// of angles.``int` `findMinimumAngle(``int` `arr[], ``int` `n)``{``    ``int` `l = 0, sum = 0, ans = 360;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// sum of array``        ``sum += arr[i];` `        ``while` `(sum >= 180) {` `            ``// calculating the difference of``            ``// angles and take minimum of``            ``// previous and newly calculated``            ``ans = min(ans, 2 * ``abs``(180 - sum));``            ``sum -= arr[l];``            ``l++;``        ``}` `        ``ans = min(ans, 2 * ``abs``(180 - sum));``    ``}``    ``return` `ans;``}` `// driver code``int` `main()``{``    ``int` `arr[] = { 100, 100, 160 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``cout << findMinimumAngle(arr, n) << endl;``    ``return` `0;``}` `// This code is contributed by "Abhishek Sharma 44"`

## Java

 `// java program to find minimum``// difference of angles of two``// parts of given circle.``import` `java.util.*;` `class` `Count{``    ``public` `static` `int` `findMinimumAngle(``int` `arr[], ``int` `n)``    ``{``        ``int` `l = ``0``, sum = ``0``, ans = ``360``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``// sum of array``            ``sum += arr[i];``    ` `            ``while` `(sum >= ``180``)``            ``{``    ` `                ``// calculating the difference of``                ``// angles and take minimum of``                ``// previous and newly calculated``                ``ans = Math.min(ans,``                            ``2` `* Math.abs(``180` `- sum));``                ``sum -= arr[l];``                ``l++;``            ``}``    ` `            ``ans = Math.min(ans,``                            ``2` `* Math.abs(``180` `- sum));``        ``}``        ` `        ``return` `ans;``        ` `    ``}``    ` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``100``, ``100``, ``160` `};``        ``int` `n = ``3``;``        ``System.out.print(findMinimumAngle(arr, n));``    ``}``}` `// This code is contributed by rishabh_jain`

## Python3

 `#Python3 program to find minimum``# difference of angles of two``# parts of given circle.``import` `math` `# function returns the minimum``# difference of angles.``def` `findMinimumAngle (arr, n):``    ``l ``=` `0``    ``_sum ``=` `0``    ``ans ``=` `360``    ``for` `i ``in` `range``(n):``        ` `        ``#sum of array``        ``_sum ``+``=` `arr[i]``        ` `        ``while` `_sum >``=` `180``:``        ` `            ``# calculating the difference of``            ``# angles and take minimum of``            ``# previous and newly calculated``            ``ans ``=` `min``(ans, ``2` `*` `abs``(``180` `-` `_sum))``            ``_sum ``-``=` `arr[l]``            ``l``+``=``1``        ``ans ``=` `min``(ans, ``2` `*` `abs``(``180` `-` `_sum))``    ``return` `ans``    ` `# driver code``arr ``=` `[``100``, ``100``, ``160``]``n ``=` `len``(arr)``print``(findMinimumAngle (arr, n))` `# This code is contributed by "Abhishek Sharma 44"`

## C#

 `// C# program to find minimum``// difference of angles of two``// parts of given circle.``using` `System;` `class` `GFG``{``    ``public` `static` `int` `findMinimumAngle(``int` `[]arr, ``int` `n)``    ``{``        ``int` `l = 0, sum = 0, ans = 360;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``// sum of array``            ``sum += arr[i];``    ` `            ``while` `(sum >= 180)``            ``{``    ` `                ``// calculating the difference of``                ``// angles and take minimum of``                ``// previous and newly calculated``                ``ans = Math.Min(ans,``                      ``2 * Math.Abs(180 - sum));``                ``sum -= arr[l];``                ``l++;``            ``}``    ` `            ``ans = Math.Min(ans,``                        ``2 * Math.Abs(180 - sum));``        ``}``        ` `        ``return` `ans;``        ` `    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = { 100, 100, 160 };``        ``int` `n = 3;``        ``Console.WriteLine(findMinimumAngle(arr, n));``    ``}``}` `// This code is contributed by vt_m`

## PHP

 `= 180)``        ``{` `            ``// calculating the difference of``            ``// angles and take minimum of``            ``// previous and newly calculated``            ``\$ans` `= min(``\$ans``, 2 *``                             ``abs``(180 - ``\$sum``));``            ``\$sum` `-= ``\$arr``[``\$l``];``            ``\$l``++;``        ``}` `        ``\$ans` `= min(``\$ans``, 2 * ``abs``(180 - ``\$sum``));``    ``}``    ``return` `\$ans``;``}` `// Driver Code``\$arr` `= ``array``( 100, 100, 160 );``\$n` `= sizeof(``\$arr``);``echo` `findMinimumAngle(``\$arr``, ``\$n``), ``"\n"` `;` `// This code is contributed by m_kit``?>`

## Javascript

 ``

Output :

`40`

Time Complexity: O(n)
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.