3024.Type_of_Triangle(Easy).md

3024. Type of Triangle (Easy)

Date and Time: Oct 23, 2024, 20:51 (EST)

Link: https://leetcode.com/problems/type-of-triangle/

Question:

You are given a 0-indexed integer array nums of size 3 which can form the sides of a triangle.

• A triangle is called equilateral if it has all sides of equal length.

• A triangle is called isosceles if it has exactly two sides of equal length.

• A triangle is called scalene if all its sides are of different lengths.

Return a string representing the type of triangle that can be formed or "none" if it cannot form a triangle.

Example 1:

Input: nums = [3,3,3]

Output: "equilateral"

Explanation: Since all the sides are of equal length, therefore, it will form an equilateral triangle.

Example 2:

Input: nums = [3,4,5]

Output: "scalene"

Explanation:
nums[0] + nums[1] = 3 + 4 = 7, which is greater than nums[2] = 5.
nums[0] + nums[2] = 3 + 5 = 8, which is greater than nums[1] = 4.
nums[1] + nums[2] = 4 + 5 = 9, which is greater than nums[0] = 3.
Since the sum of the two sides is greater than the third side for all three cases, therefore, it can form a triangle.
As all the sides are of different lengths, it will form a scalene triangle.

Constraints:

• nums.length == 3

• 1 <= nums[i] <= 100

Walk-through:

First identify if the triangle if valid or not. Then, we can uset set() to save all three sides and check how many distinct elements in set() to return the corresponding shape.

Python Brute Force:

class Solution:
    def triangleType(self, nums: List[int]) -> str:
        # First check if nums is a valid triangle
        # Valid triangle: the sum of two sides > the third side
        # Three types of triangles
        # Brute Force:
        # 1. Compare with three sides
        # 2. Check three cases

        # TC: O(1), SC: O(1)
        if nums[0] + nums[1] > nums[2] and nums[1] + nums[2] > nums[0] and nums[0] + nums[2] > nums[1]:
            if nums[0] == nums[1] == nums[2]:
                return "equilateral"
            elif nums[0] == nums[1] or nums[0] == nums[2] or nums[1] == nums[2]:
                return "isosceles"
            elif nums[0] != nums[1] and nums[1] != nums[2] and nums[0] != nums[2]:
                return "scalene"

        return "none"

Time Complexity: $O(1)$, accessing element by index takes only $O(1)$.
Space Complexity: $O(1)$

Python set Solution:

class Solution:
    def triangleType(self, nums: List[int]) -> str:
        # First check if nums is a valid triangle
        # Valid triangle: the sum of two sides > the third side
        # Three types of triangles
        # Return type by checking how many distinct elements we have

        # TC: O(1), SC: O(1)
        if nums[0] + nums[1] > nums[2] and nums[0] + nums[2] > nums[1] and nums[1] + nums[2] > nums[0]:
            nums = set(nums)
            if len(nums) == 1:
                return "equilateral"
            elif len(nums) == 2:
                return "isosceles"
            elif len(nums) == 3:
                return "scalene"
        
        return "none"

Time Complexity: $O(1)$
Space Complexity: $O(1)$, set() size is constant.