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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

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