LeetCode #3024 — EASY

Type of Triangle

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed integer array nums of size 3 which can form the sides of a triangle.

A triangle is called equilateral if it has all sides of equal length.
A triangle is called isosceles if it has exactly two sides of equal length.
A triangle is called scalene if all its sides are of different lengths.

Return a string representing the type of triangle that can be formed or "none" if it cannot form a triangle.

Example 1:

Input: nums = [3,3,3]
Output: "equilateral"
Explanation: Since all the sides are of equal length, therefore, it will form an equilateral triangle.

Example 2:

Input: nums = [3,4,5]
Output: "scalene"
Explanation: 
nums[0] + nums[1] = 3 + 4 = 7, which is greater than nums[2] = 5.
nums[0] + nums[2] = 3 + 5 = 8, which is greater than nums[1] = 4.
nums[1] + nums[2] = 4 + 5 = 9, which is greater than nums[0] = 3. 
Since the sum of the two sides is greater than the third side for all three cases, therefore, it can form a triangle.
As all the sides are of different lengths, it will form a scalene triangle.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array nums of size 3 which can form the sides of a triangle. A triangle is called equilateral if it has all sides of equal length. A triangle is called isosceles if it has exactly two sides of equal length. A triangle is called scalene if all its sides are of different lengths. Return a string representing the type of triangle that can be formed or "none" if it cannot form a triangle.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[3,3,3]

Example 2

[3,4,5]

Step 02

Core Insight

What unlocks the optimal approach

The condition for a valid triangle is that for any two sides, the sum of their lengths must be greater than the third side.
Simply count the number of unique edge lengths after checking it’s a valid triangle.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3024: Type of Triangle
class Solution {
    public String triangleType(int[] nums) {
        Arrays.sort(nums);
        if (nums[0] + nums[1] <= nums[2]) {
            return "none";
        }
        if (nums[0] == nums[2]) {
            return "equilateral";
        }
        if (nums[0] == nums[1] || nums[1] == nums[2]) {
            return "isosceles";
        }
        return "scalene";
    }
}

// Accepted solution for LeetCode #3024: Type of Triangle
func triangleType(nums []int) string {
	sort.Ints(nums)
	if nums[0]+nums[1] <= nums[2] {
		return "none"
	}
	if nums[0] == nums[2] {
		return "equilateral"
	}
	if nums[0] == nums[1] || nums[1] == nums[2] {
		return "isosceles"
	}
	return "scalene"
}

# Accepted solution for LeetCode #3024: Type of Triangle
class Solution:
    def triangleType(self, nums: List[int]) -> str:
        nums.sort()
        if nums[0] + nums[1] <= nums[2]:
            return "none"
        if nums[0] == nums[2]:
            return "equilateral"
        if nums[0] == nums[1] or nums[1] == nums[2]:
            return "isosceles"
        return "scalene"

// Accepted solution for LeetCode #3024: Type of Triangle
/**
 * [3024] Type of Triangle
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn triangle_type(nums: Vec<i32>) -> String {
        if nums[0] + nums[1] <= nums[2]
            || nums[0] + nums[2] <= nums[1]
            || nums[1] + nums[2] <= nums[0]
        {
            return "none".to_string();
        }

        if nums[0] == nums[1] || nums[0] == nums[2] || nums[1] == nums[2] {
            if nums[0] == nums[1] && nums[1] == nums[2] {
                "equilateral".to_string()
            } else {
                "isosceles".to_string()
            }
        } else {
            "scalene".to_string()
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3024() {
        assert_eq!("equilateral", Solution::triangle_type(vec![3, 3, 3]));
        assert_eq!("scalene", Solution::triangle_type(vec![3, 4, 5]));
    }
}

// Accepted solution for LeetCode #3024: Type of Triangle
function triangleType(nums: number[]): string {
    nums.sort((a, b) => a - b);
    if (nums[0] + nums[1] <= nums[2]) {
        return 'none';
    }
    if (nums[0] === nums[2]) {
        return 'equilateral';
    }
    if (nums[0] === nums[1] || nums[1] === nums[2]) {
        return 'isosceles';
    }
    return 'scalene';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.