The question is:

Write a boolean method, that get positive integer n>0. and return, if this n can be write as a power of 3's. each power, can appear at the sum maximum one time.

For example: if n = 37 f() return true, because: 3^0 3^2 3^3 = 37 if n = 38 f() return false.

I was able to solve the problem, but only when i wrote the condition (num==0) before the condition (num < 0 || 3^I > num).

can anyone explain me why?

the 2 versions of the code:

public static boolean sumPower3(int num)
    { return sumPower3(num,0); }
    // version 1(output: false):
    private static boolean sumPower3(int n,int i)
    {
        int p3i = (int)Math.pow(3,i);
        if(p3i>n || n<0){return false;}
        if(n==0){return true;}
        return sumPower3(n-p3i,i 1) || sumPower3(n,i 1);
    }
    // version 2 (output true):
    private static boolean sumPower3(int n,int i)
    {
        int p3i = (int)Math.pow(3,i);
        if(n==0){return true;}
        if(p3i>n || n<0){return false;}
        
        return sumPower3(n-p3i,i 1) || sumPower3(n,i 1);
    }

CodePudding user response：

For version 1, consider what happens in the last layer of recurision.

At this point, n is already equal to 0. The answer to this should be true, and that is what should be returned.

However, (int)Math.pow(3,i); is definitely larger than or equal to 1. This means that when n == 0, p3i>n will evaluate to true and your function returns false, giving an incorrect answer.

You should write the n == 0 condition first as p3i>n || n<0 should only return false when n != 0.

CodePudding user response：

Alternative "math" solution:

public static boolean sumPower3(int num) {
    if (num == 1 || num == 0) {
        return true;
    }
    if (num % 3 == 2) {
        return false;
    }
    return sumPower3(num / 3);
}