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326. Power of Three

Description

Given an integer n, return true if it is a power of three. Otherwise, return false.

An integer n is a power of three, if there exists an integer x such that n == 3x.

 

Example 1:

Input: n = 27
Output: true
Explanation: 27 = 33

Example 2:

Input: n = 0
Output: false
Explanation: There is no x where 3x = 0.

Example 3:

Input: n = -1
Output: false
Explanation: There is no x where 3x = (-1).

 

Constraints:

  • -231 <= n <= 231 - 1

 

Follow up: Could you solve it without loops/recursion?

Solutions

Solution 1: Trial Division

If \(n \gt 2\), we can continuously divide \(n\) by \(3\). If it's not divisible, it means \(n\) is not a power of \(3\), otherwise we continue dividing by \(3\) until \(n\) is less than or equal to \(2\). If \(n\) equals \(1\), it means \(n\) is a power of \(3\), otherwise it's not a power of \(3\).

Time complexity \(O(\log_3n)\), space complexity \(O(1)\).

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class Solution:
    def isPowerOfThree(self, n: int) -> bool:
        while n > 2:
            if n % 3:
                return False
            n //= 3
        return n == 1

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class Solution {
    public boolean isPowerOfThree(int n) {
        while (n > 2) {
            if (n % 3 != 0) {
                return false;
            }
            n /= 3;
        }
        return n == 1;
    }
}

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class Solution {
public:
    bool isPowerOfThree(int n) {
        while (n > 2) {
            if (n % 3) {
                return false;
            }
            n /= 3;
        }
        return n == 1;
    }
};

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func isPowerOfThree(n int) bool {
    for n > 2 {
        if n%3 != 0 {
            return false
        }
        n /= 3
    }
    return n == 1
}

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function isPowerOfThree(n: number): boolean {
    while (n > 2) {
        if (n % 3 !== 0) {
            return false;
        }
        n = Math.floor(n / 3);
    }
    return n === 1;
}

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impl Solution {
    pub fn is_power_of_three(mut n: i32) -> bool {
        while n > 2 {
            if n % 3 != 0 {
                return false;
            }
            n /= 3;
        }
        n == 1
    }
}

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/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfThree = function (n) {
    while (n > 2) {
        if (n % 3 !== 0) {
            return false;
        }
        n = Math.floor(n / 3);
    }
    return n === 1;
};

Solution 2: Mathematics

If \(n\) is a power of \(3\), then the maximum value of \(n\) is \(3^{19} = 1162261467\). Therefore, we only need to check if \(n\) is a divisor of \(3^{19}\).

Time complexity \(O(1)\), space complexity \(O(1)\).

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class Solution:
    def isPowerOfThree(self, n: int) -> bool:
        return n > 0 and 1162261467 % n == 0

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class Solution {
    public boolean isPowerOfThree(int n) {
        return n > 0 && 1162261467 % n == 0;
    }
}

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class Solution {
public:
    bool isPowerOfThree(int n) {
        return n > 0 && 1162261467 % n == 0;
    }
};

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func isPowerOfThree(n int) bool {
    return n > 0 && 1162261467%n == 0
}

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function isPowerOfThree(n: number): boolean {
    return n > 0 && 1162261467 % n == 0;
}

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impl Solution {
    pub fn is_power_of_three(mut n: i32) -> bool {
        n > 0 && 1162261467 % n == 0
    }
}

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/**
 * @param {number} n
 * @return {boolean}
 */
var isPowerOfThree = function (n) {
    return n > 0 && 1162261467 % n == 0;
};

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