1068EasyArrayMathSortingPolygons

Naming the Shape of Three Rods

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Three rod lengths are given as rods.

Return the name of the triangle they form:

"none" when they cannot form a triangle at all, which happens exactly when some two of them added together do not exceed the third;
"equilateral" when all three are the same length;
"isosceles" when exactly two are the same length;
"scalene" when no two are the same length.

Examples

Example 1

Input: rods = [2, 2, 3]
Output: "isosceles"

Two and two added together exceed three, so a triangle is possible, and exactly two of the rods are the same length.

Example 2

Input: rods = [1, 1, 5]
Output: "none"

The two short rods add to two, which does not exceed five, so they cannot close into a triangle at all.

Example 3

Input: rods = [6, 7, 8]
Output: "scalene"

Six and seven exceed eight, so a triangle is possible, and no two rods are the same length.

Constraints

rods.length == 3
1 <= rods[i] <= 100

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def triangle_type(rods: list[int]) -> str:

Java

public String triangleType(int[] rods)

The intended approach

Time O(1).Space O(1).

Sort the three lengths first. Then only one of the three additions needs checking: the two shorter rods added together must exceed the longest, since the other two additions each include the longest rod and so exceed the remaining one automatically. Once a triangle is possible, the name follows from how many lengths are alike. All three the same gives the first name, exactly two the same the second, and none alike the third. With the lengths sorted, "exactly two the same" is just a matter of comparing neighbours. Three values, so this is a fixed amount of work.

Pythonentry point: triangle_type

def triangle_type(rods):
    order = sorted(rods)
    if order[0] + order[1] <= order[2]:
        return "none"
    if order[0] == order[2]:
        return "equilateral"
    if order[0] == order[1] or order[1] == order[2]:
        return "isosceles"
    return "scalene"

The obvious version

Check all three additions rather than only the binding one, then count how many of the three pairs of rods happen to be the same length: three such pairs means every rod alike, one such pair means exactly two alike, and none means all different. Costs the same fixed amount but leans on no ordering.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_triangle_type(rods):
    a = rods[0]
    b = rods[1]
    c = rods[2]
    if a + b <= c or a + c <= b or b + c <= a:
        return "none"
    alike = 0
    if a == b:
        alike += 1
    if a == c:
        alike += 1
    if b == c:
        alike += 1
    if alike == 3:
        return "equilateral"
    if alike == 1:
        return "isosceles"
    return "scalene"

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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