[Leetcode][1780. Check if Number is a Sum of Powers of Three] 3 Solutions: Tricky, Recursive and Base 3 convertion

By Long Luo

This article is the solution of Problem 1780. Check if Number is a Sum of Powers of Three.

Brute Froce

This method is tricky.

We can easy get a table that recorded all the numbers using Math.power(3, n) between 11 to 10710^7.

Then scanning from the largest to the smallest, if find a value smaller or equal to nn, then subtract it from nn.

Repeat it until to 00.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
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)O(1)
  • Space Complexity: O(1)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.

1
2
3
4
5
6
7
8
9
10
public static boolean checkPowersOfThree_base3(int n) {
while (n > 0) {
if (n % 3 == 2) {
return false;
}
n /= 3;
}

return true;
}

Analysis

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

Recursive

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

1
2
3
public static boolean checkPowersOfThree_rec(int n) {
return n < 2 || (n % 3 != 2 && checkPowersOfThree_rec(n / 3));
}

Analysis

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

All suggestions are welcome.
If you have any query or suggestion please comment below.
Please upvote👍 if you like💗 it. Thank you:-)

Explore More Leetcode Solutions. 😉😃💗