Skip to content
Related Articles

Related Articles

Minimum cuts required to divide the Circle into equal parts
  • Last Updated : 16 Sep, 2019

Given an array arr which represents the different angles at which a circle is cut, the task is to determine the minimum number of more cuts required so that the circle is divided into equal parts.
Note: The array is already sorted in ascending order.

Examples:

Input: arr[] = {0, 90, 180, 270}
Output: 0
No more cuts are required as the circle is already divided into four equal parts.

Input: arr[] = {90, 210}
Output: 1
A single cut is required at 330 degree to divide the circle in three equal parts.

Approach: The idea is to calculate the Greatest Common Divisor of all the values obtained with the consecutive difference of two elements in the array in order to find the greatest (to reduce the number of cuts required) possible size for a part the circle can be divided into.



  • First store the absolute difference of 1st two values of the array in a variable named factor = arr[1] – arr[0].
  • Now traverse the array from index 2 to N-1 and for every element update factor as factor = gcd(factor, arr[i] – arr[i-1]).
  • Then for the last element update factor = gcd(factor, 360 – arr[N-1] + arr[0]).
  • Finally, the total cuts required will be (360 / factor) – N.

Below is the implementation of the above approach:

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number of cuts
// required to divide a circle into equal parts
int Parts(int Arr[], int N)
{
    int factor = Arr[1] - Arr[0];
    for (int i = 2; i < N; i++) {
        factor = __gcd(factor, Arr[i] - Arr[i - 1]);
    }
  
    // Since last part is connected with the first
    factor = __gcd(factor, 360 - Arr[N - 1] + Arr[0]);
  
    int cuts = (360 / factor) - N;
  
    return cuts;
}
  
// Driver code
int main()
{
    int Arr[] = { 0, 1 };
    int N = sizeof(Arr) / sizeof(Arr[0]);
  
    cout << Parts(Arr, N);
    return 0;
}


Java




// Java implementation of above approach
  
import java.io.*;
  
  
class GFG {
    // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0  
        if (a == 0
          return b; 
        if (b == 0
          return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    
       
  
// Function to return the number of cuts
// required to divide a circle into equal parts
static int Parts(int Arr[], int N)
{
    int factor = Arr[1] - Arr[0];
    for (int i = 2; i < N; i++) {
        factor = __gcd(factor, Arr[i] - Arr[i - 1]);
    }
  
    // Since last part is connected with the first
    factor = __gcd(factor, 360 - Arr[N - 1] + Arr[0]);
  
    int cuts = (360 / factor) - N;
  
    return cuts;
}
  
// Driver code
  
    public static void main (String[] args) {
    int Arr[] = { 0, 1 };
    int N = Arr.length;
  
    System.out.println( Parts(Arr, N));
    }
}
// This code is contributed by anuj_67..


Python 3




# Python 3 implementation of
# above approach
import math
  
# Function to return the number 
# of cuts required to divide a
# circle into equal parts
def Parts(Arr, N):
  
    factor = Arr[1] - Arr[0]
    for i in range(2, N) :
        factor = math.gcd(factor, Arr[i] - 
                                  Arr[i - 1])
      
    # Since last part is connected
    # with the first
    factor = math.gcd(factor, 360 - 
                      Arr[N - 1] + Arr[0])
  
    cuts = (360 // factor) - N
  
    return cuts
  
# Driver code
if __name__ == "__main__":
    Arr = [ 0, 1 ]
    N = len(Arr) 
  
    print( Parts(Arr, N))
  
# This code is contributed 
# by ChitraNayal


C#




//  C# implementation of above approach
  
using System;
  
class GFG
{
   // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0  
        if (a == 0) 
          return b; 
        if (b == 0) 
          return a; 
         
        // base case 
        if (a == b) 
            return a; 
         
        // a is greater 
        if (a > b) 
            return __gcd(a-b, b); 
        return __gcd(a, b-a); 
    
       
  
    // Function to return the number of cuts
    // required to divide a circle into equal parts
    static int Parts(int []Arr, int N)
    {
        int factor = Arr[1] - Arr[0];
        for (int i = 2; i < N; i++) {
            factor = __gcd(factor, Arr[i] - Arr[i - 1]);
        }
      
        // Since last part is connected with the first
        factor = __gcd(factor, 360 - Arr[N - 1] + Arr[0]);
      
        int cuts = (360 / factor) - N;
      
        return cuts;
    }
  
    // Driver code
    static void Main()
    {
            int []Arr = { 0, 1 };
            int N = Arr.Length;
            Console.WriteLine(Parts(Arr, N));
    }
}
// This code is contributed by ANKITRAI1


PHP




<?php
// PHP implementation of above approach 
  
// Recursive function to return
// gcd of a and b 
function __gcd( $a, $b
    // Everything divides 0 
    if ($a == 0) 
        return $b
    if ($b == 0) 
        return $a
      
    // base case 
    if ($a == $b
        return $a
      
    // a is greater 
    if ($a > $b
        return __gcd($a - $b, $b); 
    return __gcd($a, $b - $a); 
  
// Function to return the number of cuts 
function Parts($Arr, $N
    $factor = $Arr[1] - $Arr[0]; 
    for ($i = 2; $i < $N; $i++) 
    
        $factor = __gcd($factor, $Arr[$i] - 
                                 $Arr[$i - 1]); 
    
  
    // Since last part is connected
    // with the first 
    $factor = __gcd($factor, 360 - 
                    $Arr[$N - 1] + $Arr[0]); 
  
    $cuts = (360 / $factor) - $N
  
    return $cuts
  
// Driver code 
$Arr = array( 0, 1 ); 
$N = sizeof($Arr); 
echo (Parts($Arr, $N));
  
// This code is contributed by ajit.
?>


Output:

358

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :