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 <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" |
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
<?php // PHP program to find minimum // difference of angles of two // parts of given circle. // Returns the minimum difference // of angles. function findMinimumAngle($arr, $n) { $l = 0; $sum = 0; $ans = 360; for ($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 $arr = array( 100, 100, 160 ); $n = sizeof($arr); echo findMinimumAngle($arr, $n), "\n" ; // This code is contributed by m_kit ?> |
Javascript
<script> // javascript program to find minimum // difference of angles of two // parts of given circle. let MAXN = 109; function findMinimumAngle(arr, n) { let l = 0, sum = 0, ans = 360; for (let 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 let arr = [ 100, 100, 160 ]; let n = 3; document.write(findMinimumAngle(arr, n)); </script> |
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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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