## Find the set of Pythagorean triples from the given array

Pythagorean triplets condition: a2 + b2 = c2

### Example

Input array: [3,4,6,5,7,8]

Pythagorean triplets: 3, 4, 5

32 + 42 = 52

## Method 1

We use a brute force algorithm here :

## Algorithm

Time complexity: O(n^3)

**Step 1 :** We use 3 loops such that we take set of 3 different elements from the array.

a. We run 3 for loops. Such that for every a we take all the values b except itself. For every b we take all the values of a.

b. a, b, c are elements from the array.

**Step 2 :** we run the Pythagorean condition that is a*a + b*b = c*c, for all the sets of (a, b, c). we print them when it is true.

a. If a^2 + b^2 = c^2, print a, b, c.

### Algorithm working

## C++ Program

```
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr[] = { 3,8,4,10,6,5,12,13,27}; // array
int N = sizeof(arr)/sizeof(arr[0]);//size of array
int a,b,c;
for(int i = 0; i < N-2; i++)//select an element
{
for(int j=i+1;j <N-1; j++)//select an element in front of the considered element
{
for(int k =i+2; k<N;k++)// this element will be one ahead of the previously selected element in the jus touter loop
{
a = arr[i];
b = arr[j];
c = arr[k];
if(a*a + b*b == c*c) // if the chosen elements satisfy the pythagoras theorem then simply print the three values.
cout << a <<" "<<b<<" "<<c<<endl;
}
}
}
return 0;
}
```

## Method 2

## Algorithm

**Time complexity: O(n^2)**

**Step 1 : ** Sort the given array first using the function sort.

**Step 2 :** Instead of storing the numbers store the square of each element to directly check the Pythagorean theorem.

**Step 3 :** Take a as the smallest side, for every a check the elements from the array which satisfy the condition (a = c – b). if they satisfy this condition they form Pythagorean triplet as they satisfy the condition

a2 + b2 = c2

a. for all elements in the array from start, store the first element as “a”.

b. store the last two elements as “b” and “c” respectively.

c. Check the condition “a = c – b”. if true print the sqrts of a, b, c as a set of Pythagorean triplets.

d. If “c – b” is greater than “a”, decrease the variable pointing at the larger element(c) so that we are checking for all “c” is this condition true or not. If “c – b” is less than a decrease the variable pointing at the smaller element so that we are checking for all b is this condition true or not.

e. continue this loop for all a`s

f. If not found any, print no triplets.

### Algorithm Working

## C++ Program

```
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr[] = { 3,8,4,10,6,5,12,13,27,117,165,19,176,169,44,113,24,145,143,51,149,52,173,181,125}; // array
int N = sizeof(arr)/sizeof(arr[0]);//size of array
int a,b,c;
sort(arr,arr+N); //sort the array
for(int i=0; i < N; i++)
arr[i] = (arr[i] * arr[i]); //store the square of each element to directly check the pythagoras theorem
for(int i=0; i<N; i++)
{
int left = N-2 , right = N-1;
a = arr[i]; // first side of the triangle
while(left > i)
{
b = arr[left];
c = arr[right];
int calculated_side = c - b; //if a*a + b*b = c*c then obviously c*c - b*b = a*a , we utilize this to check the condition
if(calculated_side == a)
{
cout << sqrt(a) << " " << sqrt(b) << " " << sqrt(c) << endl;
left++; right--;
}
else if (calculated_side > a) //if side is larger than expected then decrease the variable pointing at the larger element
right--;
else // if side is smaller than expected then decrease the variable pointing at the smaller element
left--;
}
}
return 0;
}
```