Count of acute, obtuse and right triangles with given sides - GeeksforGeeks

Count of acute, obtuse and right triangles with given sides - GeeksforGeeks
Given an array of n positive distinct integers representing lengths of lines that can form triangle. The task is to find the number of acute triangles, obtuse triangles, and right triangles separately that can be formed from the given array.

Method 1 (Simple) : A brute force can be, use three loops, one for each side. Check above three conditions if a triangle is possible from three sides.

Method 2 (Efficient): An efficient approach is to first sort the array and run two loops for side a and b (a<b). Then find the farthest point q where a + b > c. So, from b to q all triangle are possible.
Also find a farthest point p where a² + b² >= c².
Now, observe if a² + b² = c², then increment count of right triangles. Also, acute triangle will be p – b – 1 and obtuse triangle will be q – p.

void findTriangle(int a[], int n)

{

    int b[n + 2];

 

    // Finding the square of each element of array.

    for (int i = 0; i < n; i++)

        b[i] = a[i] * a[i];

 

    // Sort the sides of array and their squares.

    sort(a, a + n);

    sort(b, b + n);

 

    // x for acute triangles

    // y for right triangles

    // z for obtuse triangles

    int x=0,y=0,z=0;

    for (int i=0; i<n; i++)

    {

        int p = i+1;

        int q = i+1;

 

        for (int j=i+1; j<n; j++)

        {

            // Finding the farthest point p where

            // a^2 + b^2 >= c^2.

            while (p<n-1 && b[i]+b[j]>=b[p+1])

                p++;

 

            q = max(q, p);

 

            // Finding the farthest point q where

            // a + b > c.

            while (q<n-1 && a[i]+a[j]>a[q+1])

                q++;

 

            // If point p make right triangle.

            if (b[i]+b[j]==b[p])

            {

                // All triangle between j and p are acute

                // triangles. So add p - j - 1 in x.

                x += max(p - j - 1, 0);

 

                // Increment y by 1.

                y++;

 

                // All triangle between q and p are acute

                // triangles. So add q - p in z.

                z += q - p;

            }

 

            // If no right triangle

            else

            {

                // All triangle between j and p are acute

                // triangles. So add p - j in x.

                x += max(p - j, 0);

 

                // All triangle between q and p are acute

                // triangles. So add q - p in z.

                z += q - p;

            }

        }

    }

 

    cout << "Acute Triangle: " << x << endl;

    cout << "Right Triangle: " << y << endl;

    cout << "Obtuse Triangle: " << z << endl;

}

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