Generate Pythagorean Triplets

Given a positive integer limit, your task is to find all possible Pythagorean Triplet (a, b, c), such that a <= b <= c <= limit.

Note: A Pythagorean triplet is a set of three positive integers a, b, and c such that a² + b² = c².

Input: limit = 20
Output: 3 4 5
5 12 13
6 8 10
8 15 17
9 12 15
12 16 20
Explanation: All the triplets are arranged in the format (a, b, c), where a² + b² = c².

[Naive Approach] - Using Nested Loops - O(n ^ 3) Time and O(1) Space

The idea is to use nested loops to generate all possible combinations of a, b, and c. For each combination check if a² + b² = c², if the condition satisfies store the triplet, else move to next combination.

3 4 5 
5 12 13 
6 8 10 
8 15 17 
9 12 15 
12 16 20 

Time Complexity: O(n³), where n is representing the limit.
Space Complexity: O(1)

[Expected Approach] - Using Two Pointers - O(n ^ 2) Time and O(1) Space

The idea is to use Two Pointers to generate combinations of a & b in linear time, and check if a² + b² = c². To do so, iterate through all values of c from 1 to limit, and set a = 1, and b = c - 1. At each iteration there are three possibilities:

• if a² + b² > c²: Decrease the value of b by 1.
• if a² + b² < c²: Increase the value of a by 1.
• if a² + b² = c²: Store the triplet

3 4 5 
6 8 10 
5 12 13 
9 12 15 
8 15 17 
12 16 20 

Time Complexity: O(n²), where n is representing the triplets.
Space Complexity: O(1)

[Alternate Approach] - Using Mathematics

Note: The below given method doesn't generate all triplets smaller than a given limit. For example "9 12 15" which is a valid triplet is not printed by above method.

The idea is to use square sum relation of Pythagorean Triplet that states that addition of squares of a and b is equal to square of c. We can write these numbers in terms of m and n such that,

a = m² - n²
b = 2 * m * n
c = m² + n²
because,
a² = m⁴ + n⁴ – 2 * m² * n²
b² = 4 * m² * n²
c² = m⁴ + n⁴ + 2* m² * n²

We can see that a² + b² = c², so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets. 

3 4 5
8 6 10
5 12 13
15 8 17
12 16 20

Time complexity of this approach is O(√n) where n is the given limit. We are iterating until c <= limit ,where c = m^2 + n^2. Thus the iteration will take place approximately sqrt(limit) times.
Auxiliary space: O(1) as it is using constant space for variables
References:
https://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples