Find Pythagorean Triplets from array

April 6, 2023

Find Pythagorean Triplets in C++

Pythagorean triplets are set of three values which satisfy the pythagoras theorem of a right angled triangle. Pythagoras theorem states that the square of hypotenuse is equal to the sum of square of base and square of perpendicular. The values which satisfy the above condition are considered to be as pythagorean triplets.On this page, we will learn about how to find pythagorean triplets from array in C++.

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.

Code

Run

#include <bits/stdc++.h>
using namespace std;

int main ()
{

  int arr[] = { 3, 8, 4, 10, 6, 5, 12, 13, 27 };	
  int N = sizeof (arr) / sizeof (arr[0]);	
  int a, b, c;
  for (int i = 0; i < N - 2; i++)
    {
      for (int j = i + 1; j < N - 1; j++)
	  {
	    for (int k = i + 2; k < N; k++)
	    {
	      a = arr[i];
	      b = arr[j];
	      c = arr[k];
	      if (a * a + b * b == c * c)
		      cout << a << " " << b << " " << c << endl;
	    }
	  }
    }
  return 0;
}

Output

3 4 5
8 6 10
5 12 13

Segregate 0’s 1’s and 2’s in array

Sort elements in array by frequency

Reorder array using given indexes

Merging two sorted arrays

Minimum number of Merge Operations to make an Array Palindrome

Method 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.

Code

Run

#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 };
  int N = sizeof (arr) / sizeof (arr[0]);

  int a, b, c;
  sort (arr, arr + N);
  for (int i = 0; i < N; i++)
    arr[i] = (arr[i] * arr[i]);
  for (int i = 0; i < N; i++)
  {
      int left = N - 2, right = N - 1;
      a = arr[i];

      while (left > i)
	  {
	    b = arr[left];
	    c = arr[right];

	    int calculated_side = c - b;
	    if (calculated_side == a)
	    {
	      cout << sqrt (a) << " " << sqrt (b) << " " << sqrt (c) << endl;
	      left++;
	      right --;
	    }
	    else if (calculated_side > a)
	      right --;
	    else
	      left --;
	   }
   }
  return 0;
}

Output

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