Given an array arr[] of N elements, where each element represents an angle(in degrees) of a polygon, the task is to check whether it is possible to make an N-sided polygon with all the given angles or not. If it is possible then print Yes else print No.
Examples:
Input: N = 3, arr[] = {60, 60, 60}
Output: Yes
Explanation: There exists a triangle(i.e. a polygon) satisfying the above angles. Hence the output is Yes.Input: N = 4, arr[] = {90, 90, 90, 100}
Output: No
Explanation: There does not exist any polygon satisfying the above angles. Hence the output is No.
Approach: A N-sided polygon is only possible if the sum of all the given angles is equal to 180*(N-2). Therefore the ides is to find the sum of all the angles given in the array arr[] and if the sum is equal to 180*(N-2) then print Yes, else print No.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> using namespace std;
// Function to check if the polygon // exists or not void checkValidPolygon( int arr[], int N)
{ // Initialize a variable to
// store the sum of angles
int sum = 0;
// Loop through the array and
// calculate the sum of angles
for ( int i = 0; i < N; i++) {
sum += arr[i];
}
// Check the condition for
// an N-side polygon
if (sum == 180 * (N - 2))
cout << "Yes" ;
else
cout << "No" ;
} // Driver Code int main()
{ int N = 3;
// Given array arr[]
int arr[] = { 60, 60, 60 };
// Function Call
checkValidPolygon(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to check if the polygon // exists or not static void checkValidPolygon( int arr[], int N)
{ // Initialize a variable to
// store the sum of angles
int sum = 0 ;
// Loop through the array and
// calculate the sum of angles
for ( int i = 0 ; i < N; i++)
{
sum += arr[i];
}
// Check the condition for
// an N-side polygon
if (sum == 180 * (N - 2 ))
System.out.println( "Yes" );
else
System.out.println( "No" );
} // Driver code public static void main(String[] args)
{ int N = 3 ;
// Given array arr[]
int arr[] = { 60 , 60 , 60 };
// Function call
checkValidPolygon(arr, N);
} } // This code is contributed by offbeat |
# Python3 program for the above approach # Function to check if the polygon # exists or not def checkValidPolygon(arr, N):
# Initialize a variable to
# store the sum of angles
Sum = 0
# Loop through the array and
# calculate the sum of angles
for i in range (N):
Sum + = arr[i]
# Check the condition for
# an N-side polygon
if Sum = = 180 * (N - 2 ):
print ( "Yes" )
else :
print ( "No" )
# Driver Code N = 3
# Given array arr[] arr = [ 60 , 60 , 60 ]
# Function Call checkValidPolygon(arr, N) # This code is contributed by divyeshrabadiya07 |
// C# program for the above approach using System;
class GFG{
// Function to check if the polygon // exists or not static void checkValidPolygon( int []arr, int N)
{ // Initialize a variable to
// store the sum of angles
int sum = 0;
// Loop through the array and
// calculate the sum of angles
for ( int i = 0; i < N; i++)
{
sum += arr[i];
}
// Check the condition for
// an N-side polygon
if (sum == 180 * (N - 2))
Console.Write( "Yes" );
else
Console.Write( "No" );
} // Driver code public static void Main( string [] args)
{ int N = 3;
// Given array arr[]
int []arr = { 60, 60, 60 };
// Function call
checkValidPolygon(arr, N);
} } // This code is contributed by rutvik_56 |
<script> // Javascript program for the above approach // Function to check if the polygon // exists or not function checkValidPolygon(arr, N)
{ // Initialize a variable to
// store the sum of angles
var sum = 0;
// Loop through the array and
// calculate the sum of angles
for ( var i = 0; i < N; i++)
{
sum += arr[i];
}
// Check the condition for
// an N-side polygon
if (sum == 180 * (N - 2))
document.write( "Yes" );
else
document.write( "No" );
} // Driver code var N = 3;
// Given array arr[] var arr = [ 60, 60, 60 ];
// Function call checkValidPolygon(arr, N); // This code is contributed by Kirti </script> |
Yes
Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(1)