1780. Check if Number is a Sum of Powers of Three

Description

Given an integer n, return true if it is possible to represent n as the sum of distinct powers of three. Otherwise, return false.

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

 

Example 1:

Input: n = 12
Output: true
Explanation: 12 = 31 + 32

Example 2:

Input: n = 91
Output: true
Explanation: 91 = 30 + 32 + 34

Example 3:

Input: n = 21
Output: false

 

Constraints:

• 1 <= n <= 107

Solutions

Solution 1: Mathematical Analysis

We find that if a number $n$ can be expressed as the sum of several "different" powers of three, then in the ternary representation of $n$, each digit can only be $0$ or $1$.

Therefore, we convert $n$ to ternary and then check whether each digit is $0$ or $1$. If not, then $n$ cannot be expressed as the sum of several powers of three, and we directly return false; otherwise, $n$ can be expressed as the sum of several powers of three, and we return true.

The time complexity is $O(\log_3 n)$, and the space complexity is $O(1)$.

Python3JavaC++GoTypeScript

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

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

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

func checkPowersOfThree(n int) bool {
    for n > 0 {
        if n%3 > 1 {
            return false
        }
        n /= 3
    }
    return true
}

function checkPowersOfThree(n: number): boolean {
    while (n) {
        if (n % 3 > 1) return false;
        n = Math.floor(n / 3);
    }
    return true;
}

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