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

Submitted Code

class Solution:
    def largestPerimeter(self, nums: List[int]) -> int:
        nums.sort()                 # 오름차순 정렬
        n = len(nums)
        a, b, c = n-3, n-2, n-1     # nums[a] <= nums[b] <= nums[c]

        while a >= 0:
            if (nums[a] == nums[b] == nums[c]) or (nums[a] + nums[b] > nums[c]):
                return nums[a] + nums[b] + nums[c]
            else:
                a -= 1
                b -= 1
                c -= 1

        return 0

Runtime: 10 ms | Beats 86.88%
Memory: 18.72 MB | Beats 45.77%

가장 긴 둘레를 구해야 하기 때문에 sort()로 정렬하고 뒤에서부터 시작했다. 세 변의 길이가 같거나, 가장 긴 변의 길이가 나머지 두 변의 합보다 짧으면 삼각형을 만들 수 있다.

Other Solutions

1st

class Solution:
    def largestPerimeter(self, nums: List[int]) -> int:
       nums.sort(reverse=True)
       return next((x+y+z for x,y,z in zip(nums, nums[1:], nums[2:]) if x<y+z), 0)

time complexity: 𝑂(𝑛log𝑛)
space complexity: 𝑂(1)

next()는 처음으로 조건에 만족하는 값 x+y+z를 반환한다.