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: 2
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: 1
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 trianglesint 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 codeint 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 approachimport 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 approachfrom collections import defaultdict
# Function to return the number# of unique trianglesdef 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 codeif __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 trianglesfunction 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 codelet a = [7, 5, 8, 2, 2] let b = [6, 7, 2, 3, 4] let c = [5, 6, 9, 4, 3] let n = a.lengthconsole.log(UniqueTriangles(a, b, c, n)) // This code is contributed by phasing17 |
C#
// C# implementation of the approachusing 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)