LeetCode 0976. Largest Perimeter Triangle Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Sorting + Greedy

Suppose the three sides of the triangle are a ≤ b ≤ c. The triangle has non-zero area if and only if a + b \gt c.

We can enumerate the largest side c, then select the two largest remaining sides a and b. If a + b \gt c, a triangle with non-zero area can be formed, and its perimeter will be the largest possible; otherwise, continue to enumerate the next largest side c.

The time complexity is O(n log n), and the space complexity is O(log n), where n is the length of the array nums.

PythonJavaC++GoTypeScriptRustC

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(code-box)

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; } }(code-box)

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

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

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

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 } }(code-box)

int cmp(const void* a, const void* b) { return *(int*) b - *(int*) a; } int largestPerimeter(int* nums, int numsSize) { qsort(nums, numsSize, sizeof(int), cmp); for (int i = 2; i < numsSize; i++) { if (nums[i - 2] < nums[i - 1] + nums[i]) { return nums[i - 2] + nums[i - 1] + nums[i]; } } return 0; }(code-box)