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The first approach involves using the two-pointer technique. Given an integer c, you start with two pointers a initialized at 0 and b initialized at the square root of c. You calculate a^2 + b^2 and compare it with c. If it equals c, return true. If the sum is less than c, increment a to increase the sum. If the sum is greater than c, decrement b to decrease the sum. Continue this until a exceeds b.
Time Complexity: O(sqrt(c))
Space Complexity: O(1)
1import math
2
3def judgeSquareSum(c):
4 a = 0
5 b = int(math.sqrt(c))
6 while a <= b:
7 sum = a * a + b * b
8 if sum == c:
9 return True
10 elif sum < c:
11 a += 1
12 else:
13 b -= 1
14 return FalsePython's implementation uses math.sqrt and an iterative check using the two-pointer technique.
This approach is a brute-force method. Start with a single loop iterating from a = 0 to sqrt(c). For each value of a, calculate a^2 and check if c - a^2 is a perfect square. If it is, return true, otherwise continue. If the loop completes without finding such values, return false.
Time Complexity: O(sqrt(c) * log(c)) due to the isPerfectSquare check.
Space Complexity: O(1)
1
The Java solution includes a helper function isPerfectSquare to determine if c - a^2 is a perfect square for each a.