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++
// 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
// 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
# 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 CodeN = 3# Given array arr[] arr = [ 60, 60, 60 ] # Function Call checkValidPolygon(arr, N) # This code is contributed by divyeshrabadiya07 |
C#
// C# program for the above approachusing System;class GFG{ // Function to check if the polygon// exists or notstatic 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 codepublic 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 |
Javascript
<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 codevar 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)
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