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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.
1#include <stdio.h>
2
3void checkTriangle(int x, int y, int z) {
4 if (x + y > z && x + z > y && y + z > x) {
5 printf("%d %d %d Yes\n", x, y, z);
6 } else {
7 printf("%d %d %d No\n", x, y, z);
8 }
9}
10
11int main() {
12 int triangles[][3] = {{13, 15, 30}, {10, 20, 15}};
13 int n = sizeof(triangles) / sizeof(triangles[0]);
14
15 for (int i = 0; i < n; i++) {
16 checkTriangle(triangles[i][0], triangles[i][1], triangles[i][2]);
17 }
18
19 return 0;
20}This C program defines a function checkTriangle which checks if three given sides can form a triangle using the triangle inequality theorem. The main function initializes an array of sides, iterates through each set of sides, and calls checkTriangle for each set.
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).
1
Using Java's Arrays.sort method, this program sorts the triangle sides before checking the least condition needed for the inequality theorem. Sorting ensures we make only one effective check.