LeetCode 976 - Largest Perimeter Triangle

https://leetcode.com/problems/largest-perimeter-triangle/

Given an array A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.

If it is impossible to form any triangle of non-zero area, return 0.

Example 1:

Input: [2,1,2]
Output: 5

Example 2:

Input: [1,2,1]
Output: 0

Example 3:

Input: [3,2,3,4]
Output: 10

Example 4:

Input: [3,6,2,3]
Output: 8

Note:

1. 3 <= A.length <= 10000
2. 1 <= A[i] <= 10^6

https://leetcode.com/articles/largest-perimeter-triangle/
https://leetcode.com/problems/largest-perimeter-triangle/discuss/217988/JavaC%2B%2BPython-Sort-and-Try-Biggest

For a >= b >= c, a,b,c can form a triangle if a < b + c.

1. We sort the A
2. Try to get a triangle with 3 biggest numbers.
3. If A[n-1] < A[n-2] + A[n-3], we get a triangle.
   If A[n-1] >= A[n-2] + A[n-3] >= A[i] + A[j], we cannot get any triangle with A[n-1]
4. repeat step2 and step3 with the left numbers.

Without loss of generality, say the sidelengths of the triangle are $a \leq b \leq c$. The necessary and sufficient condition for these lengths to form a triangle of non-zero area is $a + b > c$.

Say we knew $c$ already. There is no reason not to choose the largest possible $a$ and $b$ from the array. If $a + b > c$, then it forms a triangle, otherwise it doesn't.

Algorithm

This leads to a simple algorithm: Sort the array. For any $c$ in the array, we choose the largest possible $a \leq b \leq c$: these are just the two values adjacent to $c$. If this forms a triangle, we return the answer.

  public int largestPerimeter(int[] A) {

    Arrays.sort(A);

    for (int i = A.length - 3; i >= 0; --i)

      if (A[i] + A[i + 1] > A[i + 2])

        return A[i] + A[i + 1] + A[i + 2];

    return 0;

  }