Type of Triangle

Difficulty: Easy | Acceptance: 44.10% | Paid: No Topics: Array, Math, Sorting

You are given a 0-indexed array of positive integers nums. Find the type of triangle that can be formed with the side lengths nums[0], nums[1], and nums[2].

Return:

“equilateral” if it is an equilateral triangle. “isosceles” if it is an isosceles triangle. “scalene” if it is a scalene triangle. “none” if the three sides cannot form a triangle.

• Examples
• Constraints
• Approach 1: Direct Comparison
• Approach 2: Sorting

Examples

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

3 <= nums.length <= 10
1 <= nums[i] <= 10

Approach 1: Direct Comparison

Intuition We can directly check the conditions for a valid triangle and then determine the type based on the equality of the sides. For a valid triangle, the sum of any two sides must be strictly greater than the third side.

Steps

• Extract the first three elements as a, b, and c.
• Check the triangle inequality theorem: a + b > c, a + c > b, and b + c > a. If any of these fail, return “none”.
• If a == b == c, return “equilateral”.
• If any two sides are equal (a == b or b == c or a == c), return “isosceles”.
• Otherwise, return “scalene”.

class Solution:
    def triangleType(self, nums: list[int]) -> str:
        a, b, c = nums[0], nums[1], nums[2]
        if a + b <= c or a + c <= b or b + c <= a:
            return "none"
        if a == b == c:
            return "equilateral"
        if a == b or b == c or a == c:
            return "isosceles"
        return "scalene"

Complexity

• Time: O(1) - We perform a constant number of operations on 3 elements.
• Space: O(1) - No extra space is used proportional to input size.
• Notes: This is the most efficient approach as it avoids the overhead of sorting.

Approach 2: Sorting

Intuition By sorting the three side lengths, we simplify the triangle inequality check. If the sides are sorted as x <= y <= z, the triangle is valid if and only if x + y > z. This is because x + z > y and y + z > x are always true for positive numbers.

Steps

• Extract the first three elements into a new list/array.
• Sort the list in ascending order.
• Check if sides[0] + sides[1] <= sides[2]. If true, return “none”.
• Check for equilateral: if sides[0] == sides[2], all sides are equal.
• Check for isosceles: if sides[0] == sides[1] or sides[1] == sides[2].
• Otherwise, return “scalene”.

class Solution:
    def triangleType(self, nums: list[int]) -> str:
        sides = sorted(nums[:3])
        if sides[0] + sides[1] <= sides[2]:
            return "none"
        if sides[0] == sides[2]:
            return "equilateral"
        if sides[0] == sides[1] or sides[1] == sides[2]:
            return "isosceles"
        return "scalene"

Complexity

• Time: O(1) - Sorting 3 elements is constant time.
• Space: O(1) - We store at most 3 elements.
• Notes: While still O(1), this approach has slightly higher constant factors due to sorting, but offers cleaner logic for the triangle inequality check.