Question 20 || Hacker Rank

Small Triangles, Large Triangles || Hacker Rank

You are given n triangles, explicitly, their sides a_i, b_i and c_i. Print them in a similar style however arranged by their region from the littlest one to the biggest one. It is ensured that every one of the areas are unique.

 

The most effective way to compute a region of a triangle with sides a, b and c is Heron's formula:

 

 Input Configuration

The main line of each test record contains a solitary whole number n. lines follow with three space-isolated whole numbers a_i, b_i and c_i

 

Imperatives

 

Yield Configuration

Print precisely lines. On each line print space-isolated numbers and of the relating triangle.

 

Test Info 0

3

7   24   25

5   12   13

3   4   5

Test Result 0

3   4   5

5   12   13

7   24   25

 

Clarification 0

The square of the main triangle is 84. The square of the second subsequent triangle is 30. The square of the third triangle is 6. So, the arranged request is the converse i.e., reverse one.

Source Code:

#include <stdio.h>

#include <stdlib.h>

#include <math.h>

struct triangle

{

    int a;

    int b;

    int c;

};

typedef struct triangle triangle;

double area(triangle tr) 

{

    double p = (tr.a + tr.b + tr.c) / 2.0;

    return sqrt(p * (p - tr.a) * (p - tr.b) * (p - tr.c));

}

int compare(const void *a, const void *b) 

{

    triangle *t1 = (triangle*) a;

    triangle *t2 = (triangle*) b;

    double area1 = area(*t1);

    double area2 = area(*t2);

    if (area1 > area2) 

    {

        return 1;

    } else if (area1 < area2) 

    {

        return -1;

    } else 

    {

        return 0;

    }

}

void sort_by_area(triangle* tr, int n) 

{

    qsort(tr, n, sizeof(triangle), compare);

}

int main()

{

    int n;

    scanf("%d", &n);

    triangle *tr = malloc(n * sizeof(triangle));

    for (int i = 0; i < n; i++) {

        scanf("%d%d%d", &tr[i].a, &tr[i].b, &tr[i].c);

    }

    sort_by_area(tr, n);

    for (int i = 0; i < n; i++) {

        printf("%d %d %d\n", tr[i].a, tr[i].b, tr[i].c);

    }

    return 0;

}

Output: