How do you solve Sum of Square Numbers on LeetCode?

Sum of Square Numbers (LeetCode #633) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sum of two squares can be efficiently checked using mathematical properties and two-pointer technique.

What is the brute force solution for Sum of Square Numbers?

The brute force approach for Sum of Square Numbers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of integers (a, b) to see if their squares sum up to c. This is straightforward but inefficient for larger values of c. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Square Numbers?

Sum of Square Numbers uses the following concepts: Math, Two Pointers, Binary Search. The recommended patterns to study are: Two Pointers, Mathematical Properties.

What are the interview tips for Sum of Square Numbers?

Always consider edge cases, like c = 0 or c = 1. Explain your thought process clearly; it helps the interviewer understand your reasoning. Practice recognizing patterns like two pointers or binary search in similar problems.

What are common mistakes when solving Sum of Square Numbers?

Not checking both a and b properly, leading to missed valid pairs. Assuming that only positive integers can be used.

#633

Sum of Square Numbers

Medium

MathTwo PointersBinary SearchTwo PointersMathematical Properties

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a two-pointer technique, leveraging the fact that both a and b can be derived from the square root of c. This reduces the number of checks significantly.

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Algorithm

3 steps

1Step 1: Initialize two pointers, a at 0 and b at the integer square root of c.
2Step 2: While a is less than or equal to b, calculate the sum of squares: sum = a² + b².
3Step 3: If sum equals c, return true. If sum is less than c, increment a. If sum is greater than c, decrement b.

solution.py13 lines

1import math
2
3def judgeSquareSum(c):
4    a, b = 0, int(math.sqrt(c))
5    while a <= b:
6        sum_squares = a * a + b * b
7        if sum_squares == c:
8            return True
9        elif sum_squares < c:
10            a += 1
11        else:
12            b -= 1
13    return False

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the values of a and b once, adjusting them based on the sum of their squares.

1The sum of two squares can be efficiently checked using mathematical properties and two-pointer technique.
2Understanding the relationship between squares and their roots helps optimize the search.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.