README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1700-1799/1780.Check%20if%20Number%20is%20a%20Sum%20of%20Powers%20of%20Three/README_EN.md	1505	Biweekly Contest 47 Q2	Math

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

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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 $O(1)$.

Python3

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

Java

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

C++

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

Go

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

TypeScript

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