Problem

Table: Triangles

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+-------------+------+
| Column Name | Type |
+-------------+------+
| A           | int  |
| B           | int  |
| C           | int  |
+-------------+------+
(A, B, C) is the primary key for this table.
Each row include the lengths of each of a triangle's three sides.

Write a query to find the type of triangle. Output one of the following for each row:

  • Equilateral : It’s a triangle with 3 sides of equal length.
  • Isosceles : It’s a triangle with 2 sides of equal length.
  • Scalene : It’s a triangle with 3 sides of differing lengths.
  • Not A Triangle: The given values of A, B, and C don’t form a triangle.

Return the result table inany order.

The result format is in the following example.

Examples

Example 1:

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    Input: 
    Triangles table:
    +----+----+----+
    | A  | B  | C  |
    +----+----+----+
    | 20 | 20 | 23 |
    | 20 | 20 | 20 |
    | 20 | 21 | 22 |
    | 13 | 14 | 30 |
    +----+----+----+
    Output: 
    +----------------+
    | triangle_type  | 
    +----------------+
    | Isosceles      | 
    | Equilateral    |
    | Scalene        |
    | Not A Triangle |
    +----------------+
    Explanation: 
    - Values in the first row from an Isosceles triangle, because A = B.
    - Values in the second row from an Equilateral triangle, because A = B = C.
    - Values in the third row from an Scalene triangle, because A != B != C.
    - Values in the fourth row cannot form a triangle, because the combined value of sides A and B is not larger than that of side C.

Solution

Method 1 – SQL Case Classification

Intuition

A triangle is valid if the sum of any two sides is greater than the third. Once validity is checked, we classify by the number of equal sides: three (Equilateral), two (Isosceles), or none (Scalene).

Approach

  1. For each row, check if the three sides can form a triangle using the triangle inequality.
  2. If not, output ‘Not A Triangle’.
  3. If all three sides are equal, output ‘Equilateral’.
  4. If exactly two sides are equal, output ‘Isosceles’.
  5. Otherwise, output ‘Scalene’.

Code

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SELECT
  CASE
    WHEN A + B <= C OR A + C <= B OR B + C <= A THEN 'Not A Triangle'
    WHEN A = B AND B = C THEN 'Equilateral'
    WHEN A = B OR B = C OR A = C THEN 'Isosceles'
    ELSE 'Scalene'
  END AS triangle_type
FROM Triangles;
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SELECT
  CASE
    WHEN A + B <= C OR A + C <= B OR B + C <= A THEN 'Not A Triangle'
    WHEN A = B AND B = C THEN 'Equilateral'
    WHEN A = B OR B = C OR A = C THEN 'Isosceles'
    ELSE 'Scalene'
  END AS triangle_type
FROM Triangles;
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class Solution:
    def classify_triangles(self, df):
        def classify(row):
            a, b, c = row['A'], row['B'], row['C']
            if a + b <= c or a + c <= b or b + c <= a:
                return 'Not A Triangle'
            if a == b == c:
                return 'Equilateral'
            if a == b or b == c or a == c:
                return 'Isosceles'
            return 'Scalene'
        df['triangle_type'] = df.apply(classify, axis=1)
        return df[['triangle_type']]

Complexity

  • ⏰ Time complexity: O(n), where n is the number of triangles.
  • 🧺 Space complexity: O(1) (excluding input/output).