By Long Luo

This article is the solution 3 Approaches: Tricky, Recursion and Base 3 Conversion of Problem 1780. Check if Number is a Sum of Powers of Three .

Here shows 3 Approaches to slove this problem: Tricky, Recursion and Base 3 Convertion.

Brute Froce

This method is tricky.

We can easy get a table that recorded all the numbers using \(\texttt{Math.power}(3, n)\) between \(1\) to \(10^7\).

Then scanning from the largest to the smallest, if find a value smaller or equal to \(n\), then subtract it from \(n\).

Repeat it until to \(0\) .

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public static boolean checkPowersOfThree(int n) {
    int[] nums = {1, 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049, 177147, 531441, 1594323, 4782969};
    int idx = nums.length - 1;
    while (idx >= 0 && n > 0) {
        while (idx >= 0 && nums[idx] > n) {
            idx--;
        }

        if (nums[idx] == n) {
            return true;
        }

        n -= nums[idx];
        idx--;
    }

    return false;
}

Analysis

• Time Complexity: \(O(1)\)
• Space Complexity: \(O(1)\)

Base 3 Conversion

As the sum of distinct powers of \(3\), which means if we convert n to a base \(3\) , every certain position can only be \(0\) or \(1\) .

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public static boolean checkPowersOfThree_base3(int n) {
    while (n > 0) {
        if (n % 3 == 2) {
            return false;
        }
        n /= 3;
    }

    return true;
}

Analysis

• Time Complexity: \(O(\log n)\)
• Space Complexity: \(O(1)\)

Recursion

We can make our code more clean. Just One Line Is Enough!

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public static boolean checkPowersOfThree_rec(int n) {
    return n < 2 || (n % 3 != 2 && checkPowersOfThree_rec(n / 3));
}

Analysis

• Time Complexity: \(O(\log n)\)
• Space Complexity: \(O(1)\)

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