How do you solve Find Polygon With the Largest Perimeter on LeetCode?

Find Polygon With the Largest Perimeter (LeetCode #2971) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting helps to quickly identify valid sides for a polygon.

What is the time complexity of Find Polygon With the Largest Perimeter?

The optimal solution for Find Polygon With the Largest Perimeter runs in O(n log n) time and uses O(1) space. The complexity is O(n log n) due to the sorting step, which is efficient for the input size constraints.

What is the brute force solution for Find Polygon With the Largest Perimeter?

The brute force approach for Find Polygon With the Largest Perimeter is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks all combinations of three or more sides to see if they can form a polygon. This is straightforward but inefficient, as it involves checking every possible combination. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find Polygon With the Largest Perimeter?

Find Polygon With the Largest Perimeter uses the following concepts: Array, Greedy, Sorting, Prefix Sum. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Find Polygon With the Largest Perimeter?

Always consider edge cases, such as arrays with very large or very small numbers. Explain your thought process clearly while coding. Practice similar problems to recognize patterns in polygon formation.

What are common mistakes when solving Find Polygon With the Largest Perimeter?

Not checking the triangle inequality correctly. Forgetting to sort the array before checking combinations.

#2971

Find Polygon With the Largest Perimeter

Medium

ArrayGreedySortingPrefix SumGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

By sorting the array and checking the largest sides first, we can quickly determine if a valid polygon can be formed. This is efficient because it reduces the number of checks needed.

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Algorithm

4 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Iterate from the end of the sorted array to the beginning, checking sets of three consecutive sides.
3Step 3: If the longest side is less than the sum of the other two sides, return the perimeter of these three sides.
4Step 4: If no valid sides are found, return -1.

solution.py6 lines

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

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Complexity note: The complexity is O(n log n) due to the sorting step, which is efficient for the input size constraints.

1Sorting helps to quickly identify valid sides for a polygon.
2The triangle inequality principle is crucial for determining valid polygons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.