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] <= 109This approach involves sorting the array of side lengths and then iterating through it to find the largest valid polygon, specifically checking sets of three sides. By sorting, we can easily make sure that when checking three sides, the largest is the last one, allowing a simple perimeter check.
Sort the array in non-decreasing order and start checking from the third last element using a sliding window of three elements to check if they form a valid polygon (triangle) using the property: if a <= b <= c, a triangle forms if a + b > c.
In this implementation, the array is sorted in descending order. We iterate through the sorted array from the largest element and check each triplet to see if they can form a valid triangle. As soon as we find a valid triplet, we return its perimeter. If no valid polygon is found throughout the checks, we return -1.
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Time Complexity: O(n log n) due to the sorting step.
Space Complexity: O(1), as no extra space proportional to input size is used.
This approach involves checking each combination of three different side lengths from the list to see if they can form a valid polygon (triangle). For each valid combination found, we will calculate its perimeter, and keep track of the maximum perimeter obtained.
This naive method is straightforward but can be inefficient for larger lists due to its O(n^3) complexity. However, it can be insightful for understanding triangle properties in small datasets.
This brute force implementation tests every combination of three distinct sides in the sorted array. If a triplet satisfies the triangle inequality, the perimeter is calculated and compared against the current maximum to find the largest valid perimeter.
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Time Complexity: O(n^3) due to triple nested loops.
Space Complexity: O(1), apart from input storage no extra space is required.
| Approach | Complexity |
|---|---|
| Sorting and Checking Largest Triangle | Time Complexity: O(n log n) due to the sorting step. Space Complexity: O(1), as no extra space proportional to input size is used. |
| Brute Force Approach | Time Complexity: O(n^3) due to triple nested loops. Space Complexity: O(1), apart from input storage no extra space is required. |
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