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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.
1using System;
2
3class TriangleCheck {
4 static void CheckTriangle(int x, int y, int z) {
5 string result = (x + y > z && x + z > y && y + z > x) ? "Yes" : "No";
6 Console.WriteLine($"{x} {y} {z} {result}");
7 }
8
9 static void Main() {
10 int[][] triangles = new int[][] {new int[] {13, 15, 30}, new int[] {10, 20, 15}};
11 foreach (var sides in triangles) {
12 CheckTriangle(sides[0], sides[1], sides[2]);
13 }
14 }
15}This C# program defines a static method CheckTriangle that uses a conditional check to determine if three sides can form a triangle and outputs the result.
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).
This JavaScript code uses Array.sort() to order side lengths before applying a single comparison to verify if they can form a triangle. This approach limits operations to necessary checks.