LeetCode in Kotlin
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:
3 <= n <= 1051 <= nums[i] <= 109
Solution
import java.util.Collections
import java.util.PriorityQueue
class Solution {
fun largestPerimeter(nums: IntArray): Long {
var sum = 0L
val pq = PriorityQueue(Collections.reverseOrder<Long>())
for (i in nums) {
pq.add(i.toLong())
sum = (sum + i)
}
while (pq.size >= 3) {
val curr = pq.poll()
if (sum - curr > curr) {
return sum
} else {
sum -= curr
}
}
return -1
}
}