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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

Solution 1

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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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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;
}

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