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This 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.
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.
1using System;
2
3public class Solution {
4 public int LargestPerimeter(int[] nums) {
5 Array.Sort(nums);
6 for (int i = nums.Length - 1; i >= 2; --i) {
7 if (nums[i] < nums[i - 1] + nums[i - 2]) {
8 return nums[i] + nums[i - 1] + nums[i - 2];
9 }
10 }
11 return -1;
12 }
13}This C# solution first sorts the input array in ascending order using `Array.Sort()`, and then checks for valid polygonal configurations from the highest values downward. Triplets satisfying the triangle condition are checked, and the first valid triplet’s perimeter is returned.
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.
Time Complexity: O(n^3) due to triple nested loops.
Space Complexity: O(1), apart from input storage no extra space is required.
In Java, nested `for` loops allow for every triplet in the array to be checked against the triangle inequality condition. The maximum perimeter among valid triplets is returned using `Math.max()`.