LeetCode 3899: Angles of a Triangle ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2026-04-12

Submissions: Wrong Answer → Accepted

Approach
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Triangle inequality: positive area ⇔ any two sides sum to more than the third (strictly; equality makes the points collinear with zero area).
Law of cosines: cos A = (b² + c² - a²) / (2bc), A = acos(cos A), converted to degrees.
Output in non-decreasing order.

My Attempts
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Attempt 1 computed the angles correctly but returned them in the order of a, b, c without sorting. For sides = [4, 4, 1] it returned [82.8, 82.8, 14.4] instead of [14.4, 82.8, 82.8]. Adding sorted passed.

Solution
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from math import acos, degrees

class Solution:
    def internalAngles(self, sides: List[int]) -> List[float]:
        a, b, c = sorted(sides)
        if a + b <= c:
            return []                              # zero area or not a triangle
        def angle(opp: int, x: int, y: int) -> float:
            cos_v = (x * x + y * y - opp * opp) / (2 * x * y)
            return degrees(acos(max(-1.0, min(1.0, cos_v))))   # guard float error outside [-1, 1]
        return sorted([angle(a, b, c), angle(b, a, c), angle(c, a, b)])

Sorting first means the triangle inequality only needs the two shortest sides compared with the longest.

Complexity
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Time O(1), space O(1).

Mistakes to Avoid
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Sort the output.
The triangle inequality is strict.
Floating-point error can push acos’s argument slightly outside [-1, 1]; clamping is safer.

Interview Explanation
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I sort the sides; it is a triangle only if the two shortest sum to strictly more than the longest. Then the law of cosines gives each angle, converted to degrees, and the output is sorted. Time O(1).