在给定的 N 个三角形中查找唯一三角形的数量

给定三个数组 a[]、b[] 和 c[]，由 N 个元素组成，代表N 个三角形的三边。任务是从给定的三角形中找出唯一的三角形的数量。如果三角形的所有边都与其他三角形的所有边的长度匹配，则该三角形是非唯一的。
例子：

Input: a[] = {1, 2}, b[] = {2, 3}, c[] = {3, 5} 
Output: 2 
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. 
 

方法：想法是，对于每个三角形，对它的所有边进行排序，然后将其存储在地图中，如果所有这三个边都已经存在于地图中，则将频率增加 1，否则其频率将为 1。计数最终频率为 1 的地图元素的一部分将是答案。
下面是上述方法的实现：

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 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;
}

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

输出：

1

时间复杂度： O(N * log(N))
辅助空间： O(N)

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