LeetCode-in-Java

2971. Find Polygon With the Largest Perimeter

Medium

You are given an array of positive integers nums of length n.

A polygon is a closed plane figure that has at least 3 sides. The longest side of a polygon is smaller than the sum of its other sides.

Conversely, if you have k (k >= 3) positive real numbers a1, a2, a3, …, ak where a1 <= a2 <= a3 <= ... <= ak and a1 + a2 + a3 + ... + ak-1 > ak, then there always exists a polygon with k sides whose lengths are a1, a2, a3, …, ak.

The perimeter of a polygon is the sum of lengths of its sides.

Return the largest possible perimeter of a polygon whose sides can be formed from nums, or -1 if it is not possible to create a polygon.

Example 1:

Input: nums = [5,5,5]

Output: 15

Explanation: The only possible polygon that can be made from nums has 3 sides: 5, 5, and 5. The perimeter is 5 + 5 + 5 = 15.

Example 2:

Input: nums = [1,12,1,2,5,50,3]

Output: 12

Explanation: The polygon with the largest perimeter which can be made from nums has 5 sides: 1, 1, 2, 3, and 5. The perimeter is 1 + 1 + 2 + 3 + 5 = 12. We cannot have a polygon with either 12 or 50 as the longest side because it is not possible to include 2 or more smaller sides that have a greater sum than either of them. It can be shown that the largest possible perimeter is 12.

Example 3:

Input: nums = [5,5,50]

Output: -1

Explanation: There is no possible way to form a polygon from nums, as a polygon has at least 3 sides and 50 > 5 + 5.

Constraints:

Solution

import java.util.Collections;
import java.util.PriorityQueue;

public class Solution {
    public long largestPerimeter(int[] nums) {
        long sum = 0L;
        PriorityQueue<Long> pq = new PriorityQueue<>(Collections.reverseOrder());
        for (int i : nums) {
            pq.add((long) i);
            sum = (sum + i);
        }
        while (pq.size() >= 3) {
            long curr = pq.poll();
            if (sum - curr > curr) {
                return sum;
            } else {
                sum = sum - curr;
            }
        }
        return -1;
    }
}