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Given a positive integer n, return the number of triplets (a, b, c) such that 1 <= a, b, c <= n and a^2 + b^2 = c^2.

Clarifying Questions:

  1. Range of n: What is the typical or maximum value of n we can expect? This helps us optimize our code accordingly.
  2. Constraints: Are there any specific constraints like time or space complexity limits?
  3. Triplets Order: Should triplets be distinct (i.e., (a, b, c) is the same as (b, a, c)) or do we count both?

For this problem, the primary task is to clearly identify and count the Pythagorean triplets within the given range.

Strategy:

  1. Iterate Over Possible Values: Use three nested loops to iterate over all possible values of a, b, and c since the constraints average-case scenario involves a relatively small range of numbers.
  2. Check Pythagorean Triplets: For each combination of a, b, and c, check if the condition a^2 + b^2 == c^2 holds.
  3. Count Valid Triplets: Increment a counter each time a valid triplet is found.

Code:

def countSquareSumTriples(n: int) -> int:
    count = 0
    for a in range(1, n+1):
        for b in range(1, n+1):
            c2 = a*a + b*b
            c = int(c2**0.5)
            if c*c == c2 and c <= n:
                count += 1
    return count

Explanation:

  1. Variables:
    • count is used to keep track of the number of valid triplets.
  2. Loops:
    • The outer loop iterates over a from 1 to n.
    • The middle loop iterates over b from 1 to n.
    • The variable c2 represents ( a^2 + b^2 ).
    • The variable c is the square root of c2.
  3. Checking Condition:
    • Check if c squared equals c2 (i.e., c is an integer).
    • Also, check if c is within the allowed range (1 <= c <= n).
  4. Counting Valid Triplets:
    • Increment the counter if both conditions are satisfied.

Time Complexity:

This solution is efficient given the constraints, and the use of Python’s inherent mathematical operations makes it straightforward to implement and understand.

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