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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.
1def largestPerimeter(nums):
2 nums.sort(reverse=True)
3 for i in range(len(nums) - 2):
4 if nums[i] < nums[i + 1] + nums[i + 2]:
5 return nums[i] + nums[i + 1] + nums[i + 2]
6 return -1In 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.
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.
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.