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

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
```

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

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
```

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