Skip to content
Related Articles

Related Articles

Improve Article

Minimum cuts required to divide the Circle into equal parts

  • Last Updated : 14 Apr, 2021

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:
No more cuts are required as the circle is already divided into four equal parts.
Input: arr[] = {90, 210} 
Output:
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.
?>

Javascript




<script>
 
// Javascript 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
// required to divide a circle into equal parts
function Parts(Arr, N)
{
    var factor = Arr[1] - Arr[0];
    for (var 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]);
 
    var cuts = (360 / factor) - N;
 
    return cuts;
}
 
// Driver code
var Arr = [ 0, 1 ];
var N = Arr.length;
document.write( Parts(Arr, N));
 
 
</script>
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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :