Find number of unique triangles among given N triangles

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 ``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, ``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, ``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 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)

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