Minimum number of cuts required to make circle segments equal sized

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Given an array of $N$ elements where each element in the array represents the degree ( 0 <= a[i] <= 359 ) at which there is already a cut in a circle. The task is to find the minimum number of additional cuts required to make circle segments equal sized.
Examples:

Input : arr[] = { 0, 2 }
Output : 178

Input : arr[] = { 30, 60, 180 }
Output : 9

Approach : An efficient way to solve the above problem is to find gcd of all consecutive difference in angles. This gcd is the maximum angle of one circular segment and then the number of segments will be 360/gcdObtained. But, there are already N cuts. so additional cuts will be (360/gcdObtained) – N.
Below is the implementation of the above approach:

C++

// CPP program to find the minimum number
// of additional cuts required to make
// circle segments equal sized
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
int minimumCuts(int a[], int n)
{
    // Sort the array
    sort(a, a + n);
 
    // Initial gcd value
    int gcd = a[1] - a[0];
    int s = gcd;
 
    for (int i = 2; i < n; i++) {
        gcd = __gcd(gcd, a[i] - a[i - 1]);
        s += a[i] - a[i - 1];
    }
 
    // Including the last segment
    if (360 - s > 0)
        gcd = __gcd(gcd, 360 - s);
 
    return (360 / gcd) - n;
}
 
// Driver code
int main()
{
    int arr[] = { 30, 60, 180 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << minimumCuts(arr, n);
 
    return 0;
}

Java

// Java program to find the minimum 
// number of additional cuts required
// to make circle segments equal sized
import java.util.Arrays;
 
class GFG
{
     
// Recursive function to 
// return gcd of two nos
static int findgcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return findgcd(b, a % b); 
} 
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
static int minimumCuts(int a[], int n)
{
    // Sort the array
    Arrays.sort(a);
 
    // Initial gcd value
    int gcd = a[1] - a[0];
    int s = gcd;
 
    for (int i = 2; i < n; i++) 
    {
        gcd = findgcd(gcd, a[i] - a[i - 1]);
        s += a[i] - a[i - 1];
    }
 
    // Including the last segment
    if (360 - s > 0)
        gcd = findgcd(gcd, 360 - s);
 
    return (360 / gcd) - n;
}
 
// Driver code
public static void main(String[] args)
{
    int[] arr = new int[] { 30, 60, 180 };
 
    int n = arr.length;
 
    System.out.println(minimumCuts(arr, n));
}
}
 
// This code is contributed by mits

Python 3

# Python 3 program to find the minimum number
# of additional cuts required to make
# circle segments equal sized
  
import math
# Function to find the minimum number
# of additional cuts required to make
# circle segments are equal sized
def minimumCuts(a, n):
     
    # Sort the array
    a.sort()
  
    # Initial gcd value
    gcd = a[1] - a[0]
    s = gcd
  
    for i in range(2,n) :
        gcd = math.gcd(gcd, a[i] - a[i - 1])
        s += a[i] - a[i - 1]
  
    # Including the last segment
    if (360 - s > 0):
        gcd = math.gcd(gcd, 360 - s)
  
    return (360 // gcd) - n
  
# Driver code
if __name__ == "__main__":
    arr = [ 30, 60, 180 ]
  
    n = len(arr)
  
    print(minimumCuts(arr, n))

C#

// C# program to find the minimum
// number of additional cuts required
// to make circle segments equal sized
 
using System;
class GFG
{
// Recursive function to
// return gcd of two nos
static int findgcd(int a, int b)
{
if (b == 0)
return a;
 
return findgcd(b, a % b);
}
 
// Function to find the minimum number
// of additional cuts required to make
// circle segments are equal sized
static int minimumCuts(int []a, int n)
{
// Sort the array
Array.Sort(a);
 
// Initial gcd value
int gcd = a[1] - a[0];
int s = gcd;
 
for (int i = 2; i < n; i++)
{
gcd = findgcd(gcd, a[i] - a[i - 1]);
s += a[i] - a[i - 1];
}
 
// Including the last segment
if (360 - s > 0)
gcd = findgcd(gcd, 360 - s);
 
return (360 / gcd) - n;
}
 
// Driver Code
static void Main()
{
int[] arr = new int[] { 30, 60, 180 };
int n = arr.Length;
 
Console.WriteLine(minimumCuts(arr, n));
}
// This code is contributed by ANKITRAI1
}

PHP

<?php
// PHP program to find the minimum 
// number of additional cuts required 
// to make circle segments equal sized 
 
// Recursive function to return
// gcd of two nos 
function findgcd($a, $b) 
{ 
    if ($b == 0) 
        return $a; 
    return findgcd($b, $a % $b); 
} 
 
// Function to find the minimum number 
// of additional cuts required to make 
// circle segments are equal sized 
function minimumCuts($a, $n) 
{ 
    // Sort the array 
    sort($a); 
 
    // Initial gcd value 
    $gcd = $a[1] - $a[0]; 
    $s = $gcd; 
 
    for ($i = 2; $i < $n; $i++) 
    { 
        $gcd = findgcd($gcd, $a[$i] - $a[$i - 1]); 
        $s += $a[$i] - $a[$i - 1]; 
    } 
     
    // Including the last segment 
    if (360 - $s > 0) 
        $gcd = findgcd($gcd, 360 - $s); 
     
    return (360 / $gcd) - $n; 
} 
 
// Driver Code 
$arr = array(30, 60, 180); 
$n = sizeof($arr); 
 
echo (minimumCuts($arr, $n)); 
 
// This code is contributed by ajit
?>

Javascript

<script>
// javascript program to find the minimum 
// number of additional cuts required
// to make circle segments equal sized
 
    // Recursive function to
    // return gcd of two nos
    function findgcd(a , b) 
    {
        if (b == 0)
            return a;
        return findgcd(b, a % b);
    }
 
    // Function to find the minimum number
    // of additional cuts required to make
    // circle segments are equal sized
    function minimumCuts(a, n)
    {
     
        // Sort the array
        a.sort();
 
        // Initial gcd value
        var gcd = a[1] - a[0];
        var s = gcd;
 
        for (i = 2; i < n; i++) {
            gcd = findgcd(gcd, a[i] - a[i - 1]);
            s += a[i] - a[i - 1];
        }
 
        // Including the last segment
        if (360 - s > 0)
            gcd = findgcd(gcd, 360 - s);
 
        return (360 / gcd) - n;
    }
 
    // Driver code
        var arr = [ 30, 60, 180 ];
        var n = arr.length;
        document.write(minimumCuts(arr, n));
 
// This code is contributed by aashish1995 
</script>

Output:

Time complexity: O(nlogn), since the sorting best algorithm takes (nlogn) times to sort the array.

Auxiliary Space: O(1), since no extra space has been taken.

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