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To determine whether three segments can form a triangle, we can directly apply the triangle inequality theorem. For segments with lengths x, y, and z to form a triangle, the following conditions must be met:
x + y > zx + z > yy + z > xIf all three conditions are satisfied, then the segments can form a triangle.
Time Complexity: O(1) per triangle check since we only run a constant number of comparisons.
Space Complexity: O(1) since no additional space proportional to input size is used.
1public class TriangleCheck {
2 public static void checkTriangle(int x, int y, int z) {
3 System.out.print(x + " " + y + " " + z + " ");
4 if (x + y > z && x + z > y && y + z > x) {
5 System.out.println("Yes");
6 } else {
7 System.out.println("No");
8 }
9 }
10
11 public static void main(String[] args) {
12 int[][] triangles = {{13, 15, 30}, {10, 20, 15}};
13 for (int[] t : triangles) {
14 checkTriangle(t[0], t[1], t[2]);
15 }
16 }
17}This Java program uses the checkTriangle method to determine if a set of sides can form a triangle. The main method iterates through a list of tuples representing sides and calls the method for each.
In this approach, we first sort the three sides so that we only need to check one inequality condition instead of three. Given sides x, y, and z, sort them to be a ≤ b ≤ c. Then check: a + b > c.
Sorting reduces the number of comparisons and handles the integer sides efficiently when checking validity of the triangle.
Time Complexity: O(1) since sorting three items is constant time and checking has a constant cost.
Space Complexity: O(1).
Using selection-like sorting, this C program sorts the three side lengths before checking only the most significant triangle inequality condition. The sort function performs simple comparisons and swaps.