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The idea is to sort the array in non-decreasing order and try to form the triangle starting from the largest possible sides. If the largest sides satisfy the triangle inequality, they will produce the largest perimeter. Iterate from the end of the sorted array and check for the triangle inequality condition for the last three elements. If the condition is satisfied, those three can form a triangle, otherwise keep checking the next set of three largest sides.
Time Complexity: O(n log n) due to sorting the array. Space Complexity: O(1) as no additional space is used apart from auxiliary variables.
1import java.util.Arrays;
2
3class Solution {
4 public int largestPerimeter(int[] nums) {
5 Arrays.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 0;
12 }
13}The Java solution sorts the array using `Arrays.sort()`, then iterates back from the end of the sorted array. For each set of three numbers, it checks the triangle inequality condition and returns the perimeter if the condition is met.
This approach involves checking all possible combinations of three lengths from the array to see if they form a triangle. If they do, we calculate the perimeter and update the maximum perimeter found so far. This approach is less efficient due to its combinatorial nature but demonstrates the basic principle of checking all potential solutions.
Time Complexity: O(n^3), as it evaluates all combinations of three elements. Space Complexity: O(1), requiring constant additional space.
1This Python brute force solution checks every triplet of indices to see if they can form a triangle by satisfying the triangle inequality. If they can, it potentially updates the maximum perimeter result.