LeetCode 610: Triangle Judgement Solution in Python – A Step-by-Step Guide

What Is LeetCode 610: Triangle Judgement?

In LeetCode 610: Triangle Judgement, you’re given a table triangle with columns x, y, and z, each row representing three side lengths. Your task is to determine, for each row, if those lengths can form a valid triangle by checking the triangle inequality: for sides a, b, c, it must be true that a + b > c, a + c > b, and b + c > a. The output is a list of rows with an added triangle column saying "Yes" or "No." For example, [1,2,3] fails (1+2=3, not >3), but [2,3,4] works. This problem tests your ability to apply a mathematical rule to data, often framed as an SQL query, but we’ll solve it in Python for a coding twist.

Problem Statement

• Input: A list of records (or table rows) with [x, y, z].
• Output: A list of [x, y, z, triangle] where triangle is "Yes" or "No."
• Rules:
• Valid triangle: x + y > z, x + z > y, y + z > x.
• All sides are positive integers.

Constraints

• Table has at least 1 row.
• 1 <= x, y, z <= 1000.

Examples

• Input:
``` x | y | z 1 | 2 | 3 2 | 3 | 4 ```
• Output:
``` [1, 2, 3, "No"], [2, 3, 4, "Yes"] ```
• Input:
``` x | y | z 5 | 5 | 5 ```
• Output:
``` [5, 5, 5, "Yes"] ```
• Input:
``` x | y | z 1 | 1 | 3 ```
• Output:
``` [1, 1, 3, "No"] ```

These examples show the rule in action—let’s solve it!

Understanding the Problem: Judging Triangles

To solve LeetCode 610: Triangle Judgement in Python, we need to process each row of [x, y, z], apply the triangle inequality theorem, and tag it "Yes" or "No." A brute-force approach might overcomplicate things, but this is straightforward. We’ll use:

• Best Solution (Direct Inequality Check): O(n) time, O(1) space—fast and simple.
• Alternative Solution (Sorting-Based Validation): O(n) time, O(1) space—visual but adds a sort.

Let’s start with the direct solution, breaking it down for beginners.

Best Solution: Direct Triangle Inequality Check

Why This Is the Best Solution

The direct inequality check is the top choice for LeetCode 610 because it’s efficient—O(n) time with O(1) space per row—and applies the triangle inequality theorem directly, avoiding unnecessary steps. It’s the most intuitive way to test triangle validity and aligns with the problem’s simplicity. It’s like measuring sides with a ruler and checking sums on the spot!

How It Works

Think of this as a quick geometry test:

• Step 1: Process Each Row:
• Take [x, y, z] from the input.
• Step 2: Apply Inequality:
• Check: x + y > z, x + z > y, y + z > x.
• All must be true for a triangle.
• Step 3: Tag Result:
• "Yes" if all checks pass, "No" if any fail.
• Step 4: Return List:
• Append [x, y, z, triangle] for each row.

It’s like a triangle checklist—tick all boxes or bust!

Step-by-Step Example

Example:

Copy codex | y | z
1 | 2 | 3
2 | 3 | 4

• Row 1: [1, 2, 3]:
• 1 + 2 = 3, not > 3, fails.
• Result: [1, 2, 3, "No"].
• Row 2: [2, 3, 4]:
• 2 + 3 = 5 > 4, true.
• 2 + 4 = 6 > 3, true.
• 3 + 4 = 7 > 2, true.
• Result: [2, 3, 4, "Yes"].
• Output: [[1, 2, 3, "No"], [2, 3, 4, "Yes"]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def triangleJudgement(self, triangle: List[List[int]]) -> List[List]:
        # Step 1: Initialize result list
        result = []

        # Step 2: Process each row
        for row in triangle:
            x, y, z = row

            # Step 3: Check triangle inequality
            if (x + y > z) and (x + z > y) and (y + z > x):
                triangle_type = "Yes"
            else:
                triangle_type = "No"

            # Step 4: Append result
            result.append([x, y, z, triangle_type])

        # Step 5: Return result
        return result

• Line 4: Empty list for results.
• Lines 7-8: Unpack each row into x, y, z.
• Lines 11-14: Apply inequality:
• All three conditions must hold for "Yes."
• Line 16: Append row with type.
• Time Complexity: O(n)—n rows, constant check per row.
• Space Complexity: O(1)—excluding output space.

This is like a triangle validator—quick and clean!

Alternative Solution: Sorting-Based Validation

Why an Alternative Approach?

The sorting-based approach sorts the three sides first—O(n) time with O(1) space per row (since sorting 3 elements is constant). It’s more visual, reducing the check to one condition after sorting, but adds an unnecessary step for this simple case. It’s like lining up sides to see if they fit!

How It Works

Picture this as arranging sides:

• Step 1: Sort [x, y, z] to [a, b, c] (a ≤ b ≤ c).
• Step 2: Check a + b > c.
• If true, triangle works (other inequalities hold due to sorting).
• Step 3: Tag "Yes" or "No."
• Step 4: Return with original order.

It’s like a sorted triangle test—neat but extra!

Step-by-Step Example

Example:

Copy codex | y | z
1 | 2 | 3
2 | 3 | 4

• Row 1: [1, 2, 3]:
• Sort: [1, 2, 3].
• 1 + 2 = 3, not > 3, "No".
• Result: [1, 2, 3, "No"].
• Row 2: [2, 3, 4]:
• Sort: [2, 3, 4].
• 2 + 3 = 5 > 4, "Yes".
• Result: [2, 3, 4, "Yes"].
• Output: [[1, 2, 3, "No"], [2, 3, 4, "Yes"]].

Code for Sorting Approach

Copy codeclass Solution:
    def triangleJudgement(self, triangle: List[List[int]]) -> List[List]:
        result = []

        for row in triangle:
            x, y, z = row
            sides = [x, y, z]

            # Step 1: Sort sides
            sides.sort()
            a, b, c = sides

            # Step 2: Check sorted inequality
            if a + b > c:
                triangle_type = "Yes"
            else:
                triangle_type = "No"

            # Step 3: Append with original order
            result.append([x, y, z, triangle_type])

        return result

• Lines 8-12: Sort [x, y, z].
• Lines 15-18: Check a + b > c.
• Line 20: Keep original order in result.
• Time Complexity: O(n)—sorting 3 elements is O(1) per row.
• Space Complexity: O(1)—minimal extra space.

It’s a sorted triangle check—visual but overkill!

Comparing the Two Solutions

• Direct Check (Best):
• Pros: O(n) time, O(1) space, simple and fast.
• Cons: Less visual.
• Sorting (Alternative):
• Pros: O(n) time, O(1) space, single check after sort.
• Cons: Unnecessary sorting.

Direct wins for simplicity.

Additional Examples and Edge Cases

• Input: [[1,1,1]]
• Output: [[1,1,1,"Yes"]].
• Input: [[1,1,2]]
• Output: [[1,1,2,"No"]].

Both handle these well.

Complexity Breakdown

• Direct: Time O(n), Space O(1).
• Sorting: Time O(n), Space O(1).

Direct is optimal.

Key Takeaways

• Direct: Straight inequality—smart!
• Sorting: Visual check—clear!
• Math: Geometry rules are fun.
• Python Tip: Lists shine—see Python Basics.

Final Thoughts: Judge Those Triangles

LeetCode 610: Triangle Judgement in Python is a fun geometry-meets-coding challenge. The direct check offers speed and clarity, while sorting adds a visual twist. Want more? Try LeetCode 412: Fizz Buzz or LeetCode 136: Single Number for simple logic fun. Ready to measure? Head to Solve LeetCode 610 on LeetCode and judge those triangles today!