LeetCode 633: Sum of Square Numbers

Meta Data

#

Difficulty: medium First Attempt: 2026-01-25

• Submissions: Wrong Answer ×3 → Accepted ×2

Approach

#

Find non-negative integers a, b with a^2 + b^2 = c.

Converge a from 0 and b from ⌊√c⌋:

• a^2 + b^2 > c: b is too big, b -= 1.
• a^2 + b^2 < c: a is too small, a += 1.
• Equal: found.

Nothing is missed: each step rules out “the current b is too big with every value >= a” or “the current a is too small with every value <= b”, exactly like converging pointers in Two Sum II.

My Attempts

#

All three WAs were boundary issues:

int(c ** 0.5) passed. math.isqrt is safer still, since it never touches floating point.

Solution

#

from math import isqrt

class Solution:
    def judgeSquareSum(self, c: int) -> bool:
        a, b = 0, isqrt(c)
        while a <= b:
            s = a * a + b * b
            if s == c:
                return True
            if s < c:
                a += 1
            else:
                b -= 1
        return False

Complexity

#

Time O(√c), space O(1).

Mistakes to Avoid

#

• Forgetting that 0 is a square.
• Looping while a < b and missing a == b.
• Squaring a floating-point square root and comparing exactly.

Interview Explanation

#

I converge a from 0 and b from the integer square root of c: if the sum of squares is too big, decrease b; if too small, increase a; if equal, done. The boundaries matter: 0 counts, a may equal b, and isqrt avoids floating-point error. Time O(√c).