# 976. Largest Perimeter Triangle

## Description

Given an integer array nums, return the largest perimeter of a triangle with a non-zero area, formed from three of these lengths. If it is impossible to form any triangle of a non-zero area, return 0.

Example 1:

Input: nums = [2,1,2]
Output: 5
Explanation: You can form a triangle with three side lengths: 1, 2, and 2.


Example 2:

Input: nums = [1,2,1,10]
Output: 0
Explanation:
You cannot use the side lengths 1, 1, and 2 to form a triangle.
You cannot use the side lengths 1, 1, and 10 to form a triangle.
You cannot use the side lengths 1, 2, and 10 to form a triangle.
As we cannot use any three side lengths to form a triangle of non-zero area, we return 0.


Constraints:

• 3 <= nums.length <= 104
• 1 <= nums[i] <= 106

## Solutions

• class Solution {
public int largestPerimeter(int[] nums) {
Arrays.sort(nums);
for (int i = nums.length - 1; i >= 2; --i) {
int c = nums[i - 1] + nums[i - 2];
if (c > nums[i]) {
return c + nums[i];
}
}
return 0;
}
}

• class Solution {
public:
int largestPerimeter(vector<int>& nums) {
sort(nums.begin(), nums.end());
for (int i = nums.size() - 1; i >= 2; --i) {
int c = nums[i - 1] + nums[i - 2];
if (c > nums[i]) return c + nums[i];
}
return 0;
}
};

• class Solution:
def largestPerimeter(self, nums: List[int]) -> int:
nums.sort()
for i in range(len(nums) - 1, 1, -1):
if (c := nums[i - 1] + nums[i - 2]) > nums[i]:
return c + nums[i]
return 0


• func largestPerimeter(nums []int) int {
sort.Ints(nums)
for i := len(nums) - 1; i >= 2; i-- {
c := nums[i-1] + nums[i-2]
if c > nums[i] {
return c + nums[i]
}
}
return 0
}

• function largestPerimeter(nums: number[]): number {
const n = nums.length;
nums.sort((a, b) => b - a);
for (let i = 2; i < n; i++) {
const [a, b, c] = [nums[i - 2], nums[i - 1], nums[i]];
if (a < b + c) {
return a + b + c;
}
}
return 0;
}


• impl Solution {
pub fn largest_perimeter(mut nums: Vec<i32>) -> i32 {
let n = nums.len();
nums.sort_unstable_by(|a, b| b.cmp(&a));
for i in 2..n {
let (a, b, c) = (nums[i - 2], nums[i - 1], nums[i]);
if a < b + c {
return a + b + c;
}
}
0
}
}