Problème
Classify three side lengths as invalid, equilateral, isosceles, or scalene. A valid triangle has positive sides, and its two shortest sides must sum to more than its longest side. Equal-to is invalid: it forms a straight line. Use equilateral for three equal sides, isosceles for exactly two, and scalene for three different sides.
Implement triangle_classification(sides).
Input
sides: The three side lengths in any order.
Output
Return one lowercase label: invalid, equilateral, isosceles, or scalene.
Contraintes
Exactly three integer sides, each between −1,000,000,000 and 1,000,000,000. Nonpositive sides are invalid.
Exemples
Entrée :
sides = [5, 5, 8]
Sortie attendue :
expected = "isosceles"Entrée :
std::vector<long long> sides = {5, 5, 8};
Sortie attendue :
std::string expected = "isosceles";Entrée :
long[] sides = new long[] {5, 5, 8};
Sortie attendue :
string expected = "isosceles";Explication
Two sides equal 5, and 5 + 5 > 8, so this is a valid isosceles triangle.
Entrée :
sides = [1, 2, 3]
Sortie attendue :
expected = "invalid"Entrée :
std::vector<long long> sides = {1, 2, 3};
Sortie attendue :
std::string expected = "invalid";Entrée :
long[] sides = new long[] {1, 2, 3};
Sortie attendue :
string expected = "invalid";Explication
1 + 2 equals 3. The strict inequality fails, so these lengths form a line.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
First rule out impossible triangles: nonpositive sides or a longest side that is too long.
Indice 2
Sort the sides so the triangle inequality is easy to check.
Indice 3
Then classify by how many side lengths match: three, two, or none.
Complexité attendue
O(1) time, O(1) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.