README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0633.Sum%20of%20Square%20Numbers/README_EN.md	Math Two Pointers Binary Search

633. Sum of Square Numbers

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Description

Given a non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c.

 

Example 1:

Input: c = 5
Output: true
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: c = 3
Output: false

 

Constraints:

• 0 <= c <= 231 - 1

Solutions

Solution 1: Mathematics + Two Pointers

We can use the two-pointer method to solve this problem. Define two pointers $a$ and $b$, pointing to $0$ and $\sqrt{c}$ respectively. In each step, we calculate the value of $s = a^2 + b^2$, and then compare the size of $s$ and $c$. If $s = c$, we have found two integers $a$ and $b$ such that $a^2 + b^2 = c$. If $s < c$, we increase the value of $a$ by $1$. If $s > c$, we decrease the value of $b$ by $1$. We continue this process until we find the answer, or the value of $a$ is greater than the value of $b$, and return false.

The time complexity is $O(\sqrt{c})$, where $c$ is the given non-negative integer. The space complexity is $O(1)$.

Python3

class Solution:
    def judgeSquareSum(self, c: int) -> bool:
        a, b = 0, int(sqrt(c))
        while a <= b:
            s = a**2 + b**2
            if s == c:
                return True
            if s < c:
                a += 1
            else:
                b -= 1
        return False

Java

class Solution {
    public boolean judgeSquareSum(int c) {
        long a = 0, b = (long) Math.sqrt(c);
        while (a <= b) {
            long s = a * a + b * b;
            if (s == c) {
                return true;
            }
            if (s < c) {
                ++a;
            } else {
                --b;
            }
        }
        return false;
    }
}

C++

class Solution {
public:
    bool judgeSquareSum(int c) {
        long long a = 0, b = sqrt(c);
        while (a <= b) {
            long long s = a * a + b * b;
            if (s == c) {
                return true;
            }
            if (s < c) {
                ++a;
            } else {
                --b;
            }
        }
        return false;
    }
};

Go

func judgeSquareSum(c int) bool {
	a, b := 0, int(math.Sqrt(float64(c)))
	for a <= b {
		s := a*a + b*b
		if s == c {
			return true
		}
		if s < c {
			a++
		} else {
			b--
		}
	}
	return false
}

TypeScript

function judgeSquareSum(c: number): boolean {
    let [a, b] = [0, Math.floor(Math.sqrt(c))];
    while (a <= b) {
        const s = a * a + b * b;
        if (s === c) {
            return true;
        }
        if (s < c) {
            ++a;
        } else {
            --b;
        }
    }
    return false;
}

Rust

use std::cmp::Ordering;

impl Solution {
    pub fn judge_square_sum(c: i32) -> bool {
        let mut a: i64 = 0;
        let mut b: i64 = (c as f64).sqrt() as i64;
        while a <= b {
            let s = a * a + b * b;
            match s.cmp(&(c as i64)) {
                Ordering::Equal => {
                    return true;
                }
                Ordering::Less => {
                    a += 1;
                }
                Ordering::Greater => {
                    b -= 1;
                }
            }
        }
        false
    }
}