Count number of Unique Triangles using Operator overloading

Last Updated : 02 Nov, 2023

Given N triangles along with the length of their three sides as a, b and c. The task is to count the 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.

Examples:

Input: arr[] = {{3, 1, 2}, {2, 1, 4}, {4, 5, 6}, {6, 5, 4}, {4, 5, 6}, {5, 4, 6}};
Output: 3

Input: arr[] = {{4, 5, 6}, {6, 5, 4}, {1, 2, 2}, {8, 9, 12}};
Output: 3

This problem has been solved using ordered set of STL in the previous post. Approach: We will be discussing the operator overloading based approach to solve this problem where we are going to overload the relational operator (==) of our class.

Since any two sets of sides of a triangle, say {4, 5, 6}, {6, 5, 4}, are said to be equal if each element in one set corresponds to the elements in the other. So we will be checking each element of one set with the elements of the other set and keep a count of it. If the count will be the same, both the sets can simply be considered as equal. Now we have simply compared the sets using the relational operator to find the unique number of sets.
To get the number of unique sets, we can follow an approach of comparing the current set's uniqueness with the sets ahead of it. So, if there will be k sets of the same type, only the last set would be considered to be unique.

Below is C++ implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Structure to represent a Triangle
// with three sides as a, b, c
struct Triangle {
    int a, b, c;

public:
    bool operator==(const Triangle& t) const;
};

// Function to overload relational
// operator (==)
bool Triangle::operator==(const Triangle& t) const
{
    int cnt = 0;
    if ((this->a == t.a)
        || (this->a == t.b)
        || (this->a == t.c)) {
        cnt++;
    }
    if ((this->b == t.a)
        || (this->b == t.b)
        || (this->b == t.c)) {
        cnt++;
    }
    if ((this->c == t.a)
        || (this->c == t.b)
        || (this->c == t.c)) {
        cnt++;
    }

    // If all the three elements a, b, c
    // are same, triangle is not unique
    if (cnt == 3) {
        return false;
    }

    // For unique triangle return true
    return true;
}

// Function returns the number
// of unique Triangles
int countUniqueTriangles(struct Triangle arr[],
                         int n)
{

    // Unique sets
    int uni = 0;

    for (int i = 0; i < n - 1; i++) {

        // Check on uniqueness for a
        // particular set w.r.t others
        int cnt = 0;

        for (int j = i; j < n - 1; j++) {

            // Checks if two triangles
            // are different
            if (arr[i] == arr[j + 1])
                cnt++;
        }

        // If count of unique triangles
        // is same as the number of remaining
        // triangles then, increment count
        if (cnt == n - 1 - i)
            uni++;
    }

    // Since last element that
    // remains will be unique only
    return uni + 1;
}

// Driver Code
int main()
{
    // An array of structure to
    // store sides of 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);

    // Function Call
    cout << countUniqueTriangles(arr, n);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

// Class to represent a Triangle
// with three sides as a, b, c
class Triangle {
    int a, b, c;
    // Constructor
    public Triangle(int a, int b, int c)
    {
        this.a = a;
        this.b = b;
        this.c = c;
    }

    // Function to check if two triangles are equal
    @Override public boolean equals(Object obj)
    {
        if (obj == this)
            return true;

        if (!(obj instanceof Triangle))
            return false;

        Triangle t = (Triangle)obj;

        int cnt = 0;
        if ((this.a == t.a) || (this.a == t.b)
            || (this.a == t.c)) {
            cnt++;
        }
        if ((this.b == t.a) || (this.b == t.b)
            || (this.b == t.c)) {
            cnt++;
        }
        if ((this.c == t.a) || (this.c == t.b)
            || (this.c == t.c)) {
            cnt++;
        }

        // If all the three elements a, b, c
        // are same, triangle is not unique
        if (cnt == 3) {
            return false;
        }

        // For unique triangle return true
        return true;
    }
}

// Driver Class
public class Main { // Function to count the number of
                    // unique triangles
    public static int countUniqueTriangles(Triangle[] arr)
    {
        int n = arr.length;

        // Unique sets
        int uni = 0;

        for (int i = 0; i < n - 1; i++) {

            // Check on uniqueness for a
            // particular set w.r.t others
            int cnt = 0;

            for (int j = i; j < n - 1; j++) {

                // Checks if two triangles
                // are different
                if (arr[i].equals(arr[j + 1]))
                    cnt++;
            }

            // If count of unique triangles
            // is same as the number of remaining
            // triangles then, increment count
            if (cnt == n - 1 - i)
                uni++;
        }

        // Since last element that
        // remains will be unique only
        return uni + 1;
    }

    // Main Function
    public static void main(String[] args)
    {
        // An array of objects of Triangle class to
        // store sides of Triangles
        Triangle[] arr = {
            new Triangle(3, 2, 2), new Triangle(3, 4, 5),
            new Triangle(1, 2, 2), new Triangle(2, 2, 3),
            new Triangle(5, 4, 3), new Triangle(6, 4, 5)
        };

        // Function Call
        System.out.println(countUniqueTriangles(arr));
    }
}

Python3

# Python3 program for the above approach

# Structure to represent a Triangle
# with three sides as a, b, c
class Triangle:
    
    def __init__(self, a, b, c):
        self.a = a
        self.b = b
        self.c = c
    
    # Function to overload  relational
    # operator
    def __eq__(self, t):
        cnt = 0
        if self.a in [t.a, t.b, t.c]:
            cnt += 1 
        if self.b in [t.a, t.b, t.c]:
            cnt += 1
        if self.c in [t.a, t.b, t.c]:
            cnt += 1 
        
        # If all the three elements a, b, c
        # are same, triangle is not unique
        if cnt == 3:
            return False
        
        # For unique triangle, return True
        return True


# Function returns the number
# of unique Triangles
def countUniqueTriangles(arr, n):

    # Unique sets
    uni = 0;

    for i in range(n - 1):

        # Check on uniqueness for a
        # particular set w.r.t others
        cnt = 0;

        for j in range(i, n - 1):

            # Checks if two triangles
            # are different
            if (arr[i] == arr[j + 1]):
                cnt += 1
        

        # If count of unique triangles
        # is same as the number of remaining
        # triangles then, increment count
        if (cnt == n - 1 - i):
            uni+=1;
    

    # Since last element that
    # remains will be unique only
    return uni + 1;

# Driver Code

# An array of structure to
# store sides of Triangles
arr = [ Triangle(3, 2, 2), Triangle(3, 4, 5), Triangle(1, 2, 2), Triangle(2, 2, 3), Triangle(5, 4, 3), Triangle(6, 4, 5) ] 

n = len(arr)

# Function Call
print(countUniqueTriangles(arr, n))

# This code is contributed by phasing17.

using System;

// Structure to represent a Triangle
// with three sides as a, b, c
public struct Triangle
{
    public int a, b, c;

    // Function to overload relational
    // operator (==)
    public static bool operator ==(Triangle t1, Triangle t2)
    {
        int cnt = 0;
        if ((t1.a == t2.a)
            || (t1.a == t2.b)
            || (t1.a == t2.c))
        {
            cnt++;
        }
        if ((t1.b == t2.a)
            || (t1.b == t2.b)
            || (t1.b == t2.c))
        {
            cnt++;
        }
        if ((t1.c == t2.a)
            || (t1.c == t2.b)
            || (t1.c == t2.c))
        {
            cnt++;
        }

        // If all the three elements a, b, c
        // are same, triangle is not unique
        if (cnt == 3)
        {
            return false;
        }

        // For unique triangle return true
        return true;
    }

    // Function to overload relational
    // operator (!=)
    public static bool operator !=(Triangle t1, Triangle t2)
    {
        return !(t1 == t2);
    }
}

// Class to hold the main program
public class Program
{
    // Function returns the number
    // of unique Triangles
    public static int CountUniqueTriangles(Triangle[] arr, int n)
    {
        // Unique sets
        int uni = 0;

        for (int i = 0; i < n - 1; i++)
        {
            // Check on uniqueness for a
            // particular set w.r.t others
            int cnt = 0;

            for (int j = i; j < n - 1; j++)
            {
                // Checks if two triangles
                // are different
                if (arr[i] == arr[j + 1])
                    cnt++;
            }

            // If count of unique triangles
            // is same as the number of remaining
            // triangles then, increment count
            if (cnt == n - 1 - i)
                uni++;
        }

        // Since last element that
        // remains will be unique only
        return uni + 1;
    }

    // Main function
    public static void Main()
    {
        // An array of structure to
        // store sides of Triangles
        Triangle[] arr = {
            new Triangle { a = 3, b = 2, c = 2 },
            new Triangle { a = 3, b = 4, c = 5 },
            new Triangle { a = 1, b = 2, c = 2 },
            new Triangle { a = 2, b = 2, c = 3 },
            new Triangle { a = 5, b = 4, c = 3 },
            new Triangle { a = 6, b = 4, c = 5 }
        };

        int n = arr.Length;

        // Function Call
        Console.WriteLine(CountUniqueTriangles(arr, n));
    }
}
// This code is contributed by divyansh2212

JavaScript

// Class to represent a Triangle
class Triangle {
  constructor(a, b, c) {
    this.a = a;
    this.b = b;
    this.c = c;
  }

  // Function to overload relational operator
  equals(t) {
    let cnt = 0;
    if ([t.a, t.b, t.c].includes(this.a)) {
      cnt++;
    }
    if ([t.a, t.b, t.c].includes(this.b)) {
      cnt++;
    }
    if ([t.a, t.b, t.c].includes(this.c)) {
      cnt++;
    }

    // If all the three elements a, b, c
    // are same, triangle is not unique
    if (cnt === 3) {
      return false;
    }

    // For unique triangle, return True
    return true;
  }
}

// Function to count the number of unique triangles
function countUniqueTriangles(arr, n) {
  let uni = 0;

  for (let i = 0; i < n - 1; i++) {
    let cnt = 0;

    for (let j = i; j < n - 1; j++) {
      // Checks if two triangles are different
      if (arr[i].equals(arr[j + 1])) {
        cnt++;
      }
    }

    // If count of unique triangles is same as the number of remaining triangles
    // then, increment count
    if (cnt === n - 1 - i) {
      uni++;
    }
  }

  // Since the last element that remains will be unique only
  return uni + 1;
}

// Driver Code

// An array of structure to store sides of Triangles
const arr = [
  new Triangle(3, 2, 2),
  new Triangle(3, 4, 5),
  new Triangle(1, 2, 2),
  new Triangle(2, 2, 3),
  new Triangle(5, 4, 3),
  new Triangle(6, 4, 5),
];

const n = arr.length;

// Function Call
console.log(countUniqueTriangles(arr, n));

Output

Time Complexity: O(N) Auxiliary Space: O(1)

Find number of unique triangles among given N triangles

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Count number of Unique Triangles using Operator overloading

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