# 2971. Find Polygon With the Largest Perimeter

## Description

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 <= 105
• 1 <= nums[i] <= 109

## Solutions

• class Solution {
public long largestPerimeter(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
long[] s = new long[n + 1];
for (int i = 1; i <= n; ++i) {
s[i] = s[i - 1] + nums[i - 1];
}
long ans = -1;
for (int k = 3; k <= n; ++k) {
if (s[k - 1] > nums[k - 1]) {
ans = Math.max(ans, s[k]);
}
}
return ans;
}
}

• class Solution {
public:
long long largestPerimeter(vector<int>& nums) {
sort(nums.begin(), nums.end());
int n = nums.size();
vector<long long> s(n + 1);
for (int i = 1; i <= n; ++i) {
s[i] = s[i - 1] + nums[i - 1];
}
long long ans = -1;
for (int k = 3; k <= n; ++k) {
if (s[k - 1] > nums[k - 1]) {
ans = max(ans, s[k]);
}
}
return ans;
}
};

• class Solution:
def largestPerimeter(self, nums: List[int]) -> int:
nums.sort()
s = list(accumulate(nums, initial=0))
ans = -1
for k in range(3, len(nums) + 1):
if s[k - 1] > nums[k - 1]:
ans = max(ans, s[k])
return ans


• func largestPerimeter(nums []int) int64 {
sort.Ints(nums)
n := len(nums)
s := make([]int, n+1)
for i, x := range nums {
s[i+1] = s[i] + x
}
ans := -1
for k := 3; k <= n; k++ {
if s[k-1] > nums[k-1] {
ans = max(ans, s[k])
}
}
return int64(ans)
}

• function largestPerimeter(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
const s: number[] = Array(n + 1).fill(0);
for (let i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
let ans = -1;
for (let k = 3; k <= n; ++k) {
if (s[k - 1] > nums[k - 1]) {
ans = Math.max(ans, s[k]);
}
}
return ans;
}