367. Valid Perfect Square

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Description

Given a positive integer num, return true if num is a perfect square or false otherwise.

A perfect square is an integer that is the square of an integer. In other words, it is the product of some integer with itself.

You must not use any built-in library function, such as sqrt.

 

Example 1:

Input: num = 16
Output: true
Explanation: We return true because 4 * 4 = 16 and 4 is an integer.

Example 2:

Input: num = 14
Output: false
Explanation: We return false because 3.742 * 3.742 = 14 and 3.742 is not an integer.

 

Constraints:

• 1 <= num <= 231 - 1

Solutions

Solution 1: Binary Search

We can use binary search to solve this problem. Define the left boundary $l = 1$ and the right boundary $r = num$ of the binary search, then find the smallest integer $x$ that satisfies $x^2 \geq num$ in the range $[l, r]$. Finally, if $x^2 = num$, then $num$ is a perfect square.

The time complexity is $O(\log n)$, where $n$ is the given number. The space complexity is $O(1)$.

Python3

class Solution:
    def isPerfectSquare(self, num: int) -> bool:
        l = bisect_left(range(1, num + 1), num, key=lambda x: x * x) + 1
        return l * l == num

Java

class Solution {
    public boolean isPerfectSquare(int num) {
        int l = 1, r = num;
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (1L * mid * mid >= num) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l * l == num;
    }
}

C++

class Solution {
public:
    bool isPerfectSquare(int num) {
        int l = 1, r = num;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (1LL * mid * mid >= num) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return 1LL * l * l == num;
    }
};

Go

func isPerfectSquare(num int) bool {
	l := sort.Search(num, func(i int) bool { return i*i >= num })
	return l*l == num
}

TypeScript

function isPerfectSquare(num: number): boolean {
    let [l, r] = [1, num];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (mid >= num / mid) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l * l === num;
}

Rust

impl Solution {
    pub fn is_perfect_square(num: i32) -> bool {
        let mut l = 1;
        let mut r = num as i64;
        while l < r {
            let mid = (l + r) / 2;
            if mid * mid >= (num as i64) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        l * l == (num as i64)
    }
}

Solution 2: Mathematics

Since $1 + 3 + 5 + \cdots + (2n - 1) = n^2$, we can gradually subtract $1, 3, 5, \cdots$ from $num$. If $num$ finally equals $0$, then $num$ is a perfect square.

The time complexity is $O(\sqrt n)$, and the space complexity is $O(1)$.

Python3

class Solution:
    def isPerfectSquare(self, num: int) -> bool:
        i = 1
        while num > 0:
            num -= i
            i += 2
        return num == 0

Java

class Solution {
    public boolean isPerfectSquare(int num) {
        for (int i = 1; num > 0; i += 2) {
            num -= i;
        }
        return num == 0;
    }
}

C++

class Solution {
public:
    bool isPerfectSquare(int num) {
        for (int i = 1; num > 0; i += 2) {
            num -= i;
        }
        return num == 0;
    }
};

Go

func isPerfectSquare(num int) bool {
	for i := 1; num > 0; i += 2 {
		num -= i
	}
	return num == 0
}

TypeScript

function isPerfectSquare(num: number): boolean {
    let i = 1;
    while (num > 0) {
        num -= i;
        i += 2;
    }
    return num === 0;
}

Rust

impl Solution {
    pub fn is_perfect_square(mut num: i32) -> bool {
        let mut i = 1;
        while num > 0 {
            num -= i;
            i += 2;
        }
        num == 0
    }
}