633. Sum of Square Numbers

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 $O(1)$.

Python3JavaC++GoTypeScriptRust

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

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;
    }
}

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;
    }
};

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
}

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;
}

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
    }
}

Solution 2: Mathematics

This problem is essentially about the conditions under which a number can be expressed as the sum of two squares. This theorem dates back to Fermat and Euler and is a classic result in number theory.

Specifically, the theorem can be stated as follows:

A positive integer $n$ can be expressed as the sum of two squares if and only if all prime factors of $n$ of the form $4k + 3$ have even powers.

This means that if we decompose $n$ into the product of its prime factors, $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, where $p_i$ are primes and $e_i$ are their corresponding exponents, then $n$ can be expressed as the sum of two squares if and only if all prime factors $p_i$ of the form $4k + 3$ have even exponents $e_i$.

More formally, if $p_i$ is a prime of the form $4k + 3$, then for each such $p_i$, $e_i$ must be even.

For example:

The number $13$ is a prime and $13 \equiv 1 \pmod{4}$, so it can be expressed as the sum of two squares, i.e., $13 = 2^2 + 3^2$.
The number $21$ can be decomposed into $3 \times 7$, where both $3$ and $7$ are prime factors of the form $4k + 3$ and their exponents are $1$ (odd), so $21$ cannot be expressed as the sum of two squares.

In summary, this theorem is very important in number theory for determining whether a number can be expressed as the sum of two squares.