Table of Contents
Problem Statement
We have given an array that contains n integers. We need to find the set of Pythagorean triples from the given array.
Note: Pythagorean triplets condition: a^2 + b^2 = c^2.
Example
Input
6
[3, 4, 6, 5, 7, 8]
Output
Pythagorean triplets: 3, 4, 5
Approach 1
We use a brute force algorithm here :
Algorithm
Step 1: We use 3 loops such that we take a set of 3 different elements from the array.
a. We run 3 for loops. Such that for every 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.
Implementation
C++ Program to Find Pythagorean Triplets from Array
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
cin>>N;
int arr[N];
for(int i=0;i<N;i++)
{
cin>>arr[i];
}
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;
}Java Program to Find Pythagorean Triplets from Array
import static java.lang.Integer.max;
import java.util.Scanner;
class sum
{
public static void main(String[] args)
{
Scanner sr = new Scanner(System.in);
int n = sr.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++)
{
arr[i] = sr.nextInt();
}
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.
System.out.println(a +" "+b+" "+c);
}
}
}
}
}6 3 4 6 5 7 8
3 4 5
Complexity Analysis to Find Pythagorean Triplets from Array
Time complexity
O(n^3) where n is the size of the array. Here we check the condition for every possible triplet.
Space Complexity
O(1) because we use a few variables to calculate the answer.
Approach 2
Algorithm
- Sort the given array first using the function sort.
- Instead of storing the numbers store the square of each element to directly check the Pythagorean theorem.
- 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 a^2 + b^2 = c^2
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 elements 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.
Implementation
C++ Program to Find Pythagorean Triplets from Array
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
cin>>N;
int arr[N];
for(int i=0;i<N;i++)
{
cin>>arr[i];
}
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;
}Java Program to Find Pythagorean Triplets from Array
import static java.lang.Math.sqrt;
import java.util.Arrays;
import java.util.Scanner;
class sum
{
public static void main(String[] args)
{
Scanner sr = new Scanner(System.in);
int n = sr.nextInt();
int arr[] = new int[n];
for(int i=0;i<n;i++)
{
arr[i] = sr.nextInt();
}
int a,b,c;
Arrays.sort(arr); //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)
{
System.out.println((int)sqrt(a) + " " + (int)sqrt(b) + " " + (int)sqrt(c));
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--;
}
}
}
}25 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
3 4 5 5 12 13 6 8 10 24 143 145 44 117 125 52 165 173
Complexity Analysis to Find Pythagorean Triplets from Array
Time complexity
O(n^2) where n is the size of the array. Here we use two pointer methods for each I value. This leads us to O(N*N) time complexity.
Space Complexity
O(1) because we use a few variables to calculate the answer.