We are given n triangles along with length of their three sides as a,b,c. Now we need to count number of unique triangles out of these n given triangles. Two triangles are different from one another if they have at least one of the sides different.
Example:
Input: arr[] = {{1, 2, 2},
{4, 5, 6},
{4, 5, 6}
Output: 2
Input: arr[] = {{4, 5, 6}, {6, 5, 4},
{1, 2, 2}, {8, 9, 12}
Output: 3
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Since we are asked to find number of “unique” triangles, we can use either set or unordered_set. In this post, set based approach is discussed.
How to store three sides as an element in the container? We use STL pair to store all the three sides together as
pair <int, pair<int, int> >
We one by one insert all triangles into the set. But the problem with this approach is that a triangle with sides as {4, 5, 6} is different from a triangle with sides {5, 4, 6} although they refer to a same triangle.
In order to handle such cases, we store sides in sorted order (on the basis of length of sides), here sorting won’t be an issue since we have only 3 sides and we can sort them in constant time. For example {5, 4, 6} is inserted into set as {4, 5, 6}
Note : We can make pair either by make_pair(a,b) or we can simply use {a, b}.
Below is C++ implementation of the above idea:
// A C++ program to find number of unique Triangles #include <bits/stdc++.h> using namespace std; // Creating shortcut for an integer pair. typedef pair<int, int> iPair; // A structure to represent a Triangle with // three sides as a, b, c struct Triangle { int a, b, c; }; // A function to sort three numbers a, b and c. // This function makes 'a' smallest, 'b' middle // and 'c' largest (Note that a, b and c are passed // by reference) int sort3(int &a, int &b, int &c) { vector<int> arr({a, b, c}); sort(arr.begin(), arr.end()); a = arr[0]; b = arr[1]; c = arr[2]; } // Function returns the number of unique Triangles int countUniqueTriangles(struct Triangle arr[], int n) { // A set which consists of unique Triangles set < pair< int, iPair > > s; // Insert all triangles one by one for (int i=0; i<n; i++) { // Find three sides and sort them int a = arr[i].a, b = arr[i].b, c = arr[i].c; sort3(a, b, c); // Insert a triangle into the set s.insert({a, {b, c}}); } // Return set size return s.size(); } // Driver program to test above function int main() { // An array of structure to store sides of 6 Triangles struct Triangle arr[] = {{3, 2, 2}, {3, 4, 5}, {1, 2, 2}, {2, 2, 3}, {5, 4, 3}, {6, 4, 5}}; int n = sizeof(arr)/sizeof(Triangle); cout << "Number of Unique Triangles are " << countUniqueTriangles(arr, n); return 0; } |
Output:
Number of Unique Triangles are 4
The time complexity of above solution is O(n Log n) as set requires O(Log n) time for insertion.
We can improve time complexity to O(n) using unordered_set. But using unordered_set requires writing of a hash function as hash function is not defined in library for pairs.
Related Article: Number of Possible Triangles
This article is contributed by Chirag Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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