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Find number of unique triangles among given N triangles

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Given three arrays a[], b[], and c[] of N elements representing the three sides of N triangles. The task is to find the number of triangles that are unique out of given triangles. A triangle is non-unique if all of its sides match with all the sides of some other triangle in length.
Examples:  

Input: a[] = {1, 2}, b[] = {2, 3}, c[] = {3, 5} 
Output:
Explanation:
The triangles have sides 1, 2, 3 and 2, 3, 5 respectively. 
None of them have same sides. Thus both are unique.

Input: a[] = {7, 5, 8, 2, 2}, b[] = {6, 7, 2, 3, 4}, c[] = {5, 6, 9, 4, 3} 
Output:
Only triangle with sides 8, 2 and 9 is unique. 


 


Approach: The idea is, for each triangle, sort all of its sides and then store it in a map, if all those three sides are already present in the map then increase the frequency by 1, else its frequency will be 1. The count of elements of the map which have frequency 1 in the end will be the answer.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number of unique triangles
int UniqueTriangles(int a[], int b[], int c[], int n)
{
    vector<int> sides[n];
 
    // Map to store the frequency of triangles
    // with same sides
    map<vector<int>, int> m;
 
    for (int i = 0; i < n; i++) {
 
        // Push all the sides of the current triangle
        sides[i].push_back(a[i]);
        sides[i].push_back(b[i]);
        sides[i].push_back(c[i]);
 
        // Sort the three sides
        sort(sides[i].begin(), sides[i].end());
 
        // Store the frequency of the sides
        // of the triangle
        m[sides[i]] = m[sides[i]] + 1;
    }
 
    map<vector<int>, int>::iterator i;
 
    // To store the count of unique triangles
    int count = 0;
    for (i = m.begin(); i != m.end(); i++) {
 
        // If current triangle has unique sides
        if (i->second == 1)
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    int a[] = { 7, 5, 8, 2, 2 };
    int b[] = { 6, 7, 2, 3, 4 };
    int c[] = { 5, 6, 9, 4, 3 };
 
    int n = sizeof(a) / sizeof(int);
 
    cout << UniqueTriangles(a, b, c, n);
 
    return 0;
}

                    

Java

// Java Code for above approach
import java.util.*;
 
public class Solution
{
 
  // Function to return the number
  // of unique triangles
  static int UniqueTriangles(int[] a, int[] b, int[] c,
                             int n)
  {
    String[][] sides = new String[n][3];
 
    // Map to store the frequency of
    // triangles with same sides
    HashMap<String, Integer> m = new HashMap<>();
 
    for (var i = 0; i < n; i++) {
 
      // Sort the three sides
      String[] arr
        = { a[i] + "", b[i] + "", c[i] + "" };
      Arrays.sort(arr);
 
      // Push all the sides of the current triangle
      sides[i] = arr;
 
      // Store the frequency of the sides
      // of the triangle
      String key = String.join(",", sides[i]);
      if (!m.containsKey(key))
        m.put(key, 0);
      m.put(key, m.getOrDefault(key, m.get(key)) + 1);
    }
 
    // To store the count of unique triangles
    int count = 0;
    for (String s : m.keySet())
 
      // If current triangle has unique sides
      if (m.get(s) == 1)
        count++;
 
    return count;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int[] a = { 7, 5, 8, 2, 2 };
    int[] b = { 6, 7, 2, 3, 4 };
    int[] c = { 5, 6, 9, 4, 3 };
 
    int n = a.length;
    System.out.println(UniqueTriangles(a, b, c, n));
  }
}
 
// This code is contributed by karandeep1234

                    

Python3

# Python3 implementation of the approach
from collections import defaultdict
 
# Function to return the number
# of unique triangles
 
 
def UniqueTriangles(a, b, c, n):
 
    sides = [None for i in range(n)]
 
    # Map to store the frequency of
    # triangles with same sides
    m = defaultdict(lambda: 0)
 
    for i in range(0, n):
 
        # Push all the sides of the current triangle
        sides[i] = (a[i], b[i], c[i])
 
        # Sort the three sides
        sides[i] = tuple(sorted(sides[i]))
 
        # Store the frequency of the sides
        # of the triangle
        m[sides[i]] += 1
 
    # To store the count of unique triangles
    count = 0
    for i in m:
 
        # If current triangle has unique sides
        if m[i] == 1:
            count += 1
 
    return count
 
 
# Driver code
if __name__ == "__main__":
 
    a = [7, 5, 8, 2, 2]
    b = [6, 7, 2, 3, 4]
    c = [5, 6, 9, 4, 3]
 
    n = len(a)
 
    print(UniqueTriangles(a, b, c, n))
 
# This code is contributed by Rituraj Jain

                    

Javascript

// JS implementation of the approach
 
// Function to return the number
// of unique triangles
function UniqueTriangles(a, b, c, n)
{
 
    let sides = new Array(n)
 
    // Map to store the frequency of
    // triangles with same sides
    let m = {}
     
    for (var i = 0; i < n; i++)
    {
 
        // Sort the three sides
        let arr = [a[i], b[i], c[i]]
        arr.sort()
 
        // Push all the sides of the current triangle
        sides[i] = arr
 
        // Store the frequency of the sides
        // of the triangle
        let key = sides[i].join('#')
        if (!m.hasOwnProperty(key))
            m[key] = 0
        m[key] += 1
    }
     
    // To store the count of unique triangles
    let count = 0
    for (let [i, val] of Object.entries(m))
 
        // If current triangle has unique sides
        if (m[i] == 1)
            count += 1
 
    return count
}
 
// Driver code
let a = [7, 5, 8, 2, 2]
let b = [6, 7, 2, 3, 4]
let c = [5, 6, 9, 4, 3]
 
let n = a.length
 
console.log(UniqueTriangles(a, b, c, n))
     
// This code is contributed by phasing17

                    

C#

// C# implementation of the approach
 
using System;
using System.Collections.Generic;
 
class GFG {
    // Function to return the number
    // of unique triangles
    static int UniqueTriangles(int[] a, int[] b, int[] c,
                               int n)
    {
 
        int[][] sides = new int[n][];
 
        // Map to store the frequency of
        // triangles with same sides
        Dictionary<string, int> m
            = new Dictionary<string, int>();
 
        for (var i = 0; i < n; i++) {
 
            // Sort the three sides
            int[] arr = { a[i], b[i], c[i] };
            Array.Sort(arr);
 
            // Push all the sides of the current triangle
            sides[i] = arr;
 
            // Store the frequency of the sides
            // of the triangle
            string key = string.Join(",", sides[i]);
            if (!m.ContainsKey(key))
                m[key] = 0;
            m[key] += 1;
        }
 
        // To store the count of unique triangles
        int count = 0;
        foreach(var entry in m)
 
            // If current triangle has unique sides
            if (entry.Value == 1) count++;
 
        return count;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int[] a = { 7, 5, 8, 2, 2 };
        int[] b = { 6, 7, 2, 3, 4 };
        int[] c = { 5, 6, 9, 4, 3 };
 
        int n = a.Length;
 
        Console.WriteLine(UniqueTriangles(a, b, c, n));
    }
}
 
// This code is contributed by phasing17

                    

Output: 
1

 

Time Complexity : O(N * log(N))
Auxiliary Space: O(N) 



Last Updated : 08 Nov, 2022
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