Given a division of circle into n pieces as an array of size n. The i-th 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} Ouput : 40

We can notice that if one of the part 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 it’s angle and update answer.

## C++

// CPP program to find minimum difference // of angles of two parts of given circle. #include <bits/stdc++.h> 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"

## Python3

# Python3 code 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"

Output:

40

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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.