# Small Triangles, Large Triangles in c – Hacker Rank Solution Small Triangles, Large Triangles in c – Hacker Rank Solution

You are given n, triangles, specifically, their sides ai, bi and ci. Print them in the same style but sorted by their areas from the smallest one to the largest one. It is guaranteed that all the areas are different. The best way to calculate a volume of the triangle with sides a, b and c is Heron’s formula:
s = √(P * (P – a) * (P – b) * (P – c)) where P = (a+b+c)/2.

#### Input Format

First line of each test file contains a single integer n.n lines follow with ai, bi and con each separated by single spaces.

#### Constraints

• 1<=n<=100
• 1<= ai, bi, ci <=70
• ai + bi > ci, ai + ci > bi and bi+ci > ai

#### Output Format

Print exactly n lines. On each line print 3 integers separated by single spaces, which are ai, bi and cof the corresponding triangle.

```3
7 24 25
5 12 13
3 4 5
```

```3 4 5
5 12 13
7 24 25
```

#### Explanation

The square of the first triangle is 84. The square of the second triangle is 30.The square of the third triangle is 6. So the sorted order is the reverse one.

### Small Triangles, Large Triangles in c – Hacker Rank Solution

```#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct triangle
{
int a;
int b;
int c;
};
typedef struct triangle triangle;
void sort_by_area(triangle* tr, int n)
{
int *p=malloc(n*sizeof(float));
//create array of size n to store "volumes"
for(int i=0;i<n;i++)
{
float a=(tr[i].a+tr[i].b+tr[i].c)/2.0;
//use 2.0 compulsary int/int gives int, int/float gives float
p[i]=(a*(a-tr[i].a)*(a-tr[i].b)*(a-tr[i].c));
//formula without sqrt as areas are different guarenteed
//because sqrt dosent work well with float values
}
//bubble sort
for(int i=0;i<n;i++)
{
for(int j=0;j<n-i-1;j++)
{
if(p[j]>p[j+1])
{
int temp=p[j];
p[j]=p[j+1];
p[j+1]=temp;
//swapping array of areas in ascending
//and simuntaneously the structure contents
temp=tr[j].a;
tr[j].a=tr[j+1].a;
tr[j+1].a=temp;
temp=tr[j].b;
tr[j].b=tr[j+1].b;
tr[j+1].b=temp;
temp=tr[j].c;
tr[j].c=tr[j+1].c;
tr[j+1].c=temp;
}
}
}
}
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;
}```

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