Sum of sides of largest and smallest child polygons possible from a given polygon

Sum of sides of largest and smallest child polygons possible from a given polygon
Given an integer A representing the external angle (in degrees) of a regular convex polygon, the task is to find the sum of the sides of the largest and smallest secondary polygons formed such that each edge of the secondary polygon is a chord of the primary polygon. If it is not possible to form such polygons, then print “-1”.
Examples:

Input: A = 45Output: 7Explanation:The primary polygon is an Octagon of 8 sides.Therefore, the smallest secondary polygon consists of 3 edges and the largest secondary polygon consists of 4 edges.

Sum of edges of the smallest secondary polygon + Edges of the largest secondary polygon = 3 + 4 = 7.Input: A = 60Output: 6Explanation: The primary polygon is a Hexagon consisting of 6 sides. Therefore, the smallest secondary polygon consists of 3 edges and the largest secondary polygon consists of 3 edges.

Approach: The idea is to first find the number of edges in the primary polygon and then, check whether it is possible to make secondary polygons or not. Follow the steps below to solve the problem:
The sum of the external angle in a regular polygon is 360 degrees. Therefore, the number of sides = 360 / external angle.
The number of sides of the maximum secondary polygon is the number of sides of the primary polygon / 2.
As polygon is possible if the count of edges is at least 3, a secondary polygon is possible if the edges of the initial polygon ≥ 6.
The smallest possible polygon has 3 edges always.
Print the sum of number of sides in both the largest and smallest polygons.
Below is the implementation of the above approach:

Python3

  

def secondary_polygon(Angle):
  
    
    edges_primary = 360//Angle
  
    if edges_primary >= 6:
  
        
        
        edges_max_secondary = edges_primary // 2
  
        return edges_max_secondary + 3
  
    else:
        return “Not Possible”
  
  

if __name__ == ‘__main__’:
  
    
    Angle = 45
    print(secondary_polygon(Angle))

Time Complexity: O(1)Auxiliary Space: O(1)

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