633 Sum of Square Numbers

Instead of looking for both numbers $a^2$ and $b^2$ from 0 to c with a double for loop, we can save a lot of repeated calculations by using a two-pointer strategy.

Since our required expression is a^2 + b^2 = c, we know that the maximum value either a or b can have is sqrt(c), because if any one (or both) exceeds sqrt(c), the sum of both the squares of a and b would exceed c.

Code:

Python3

def judgeSquareSum(self, c: int) -> bool:
    a, b = 0, int(sqrt(c))
 
    while a <= b:
         x = (a*a) + (b*b)
         if x == c: return True
         elif x < c: a += 1
         elif x > c: b -= 1
    
    return False

Big O Analysis

• Runtime

  The runtime complexity here is $O(sqrt(c))$ since our sample space to look for would be a - b were b = sqrt(c)

• Memory

  The memory usage is $O(1)$ since we are not using any extra data structure.

— A

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