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Question

Formatted question description: https://leetcode.ca/all/326.html

Given an integer n, return true if it is a power of three. Otherwise, return false.

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

 

Example 1:

Input: n = 27
Output: true
Explanation: 27 = 33

Example 2:

Input: n = 0
Output: false
Explanation: There is no x where 3x = 0.

Example 3:

Input: n = -1
Output: false
Explanation: There is no x where 3x = (-1).

 

Constraints:

  • -231 <= n <= 231 - 1

 

Follow up: Could you solve it without loops/recursion?

Algorithm

The most straightforward way is to keep dividing by 3 to see if the final iterative quotient is 1. Pay attention to the case where the input is negative and 0.

If no loop, there is a tricky method. Since the input is an int, the range of positive numbers is 0 - 2^31, and the largest power of 3 allowed in this range is 3^19=1162261467, so we only need to see if this number is divisible by n.

Code

  • 
    public class Power_of_Three {
    
        public class Solution_math {
            public boolean isPowerOfThree(int n) {
                return (n > 0 && 1162261467 % n == 0);
            }
        }
    
        public class Solution {
            public boolean isPowerOfThree(int n) {
                if (n < 1) {
                    return false;
                }
    
                while (n % 3 == 0) {
                    n /= 3;
                }
    
                return n == 1;
            }
        }
    }
    
    ############
    
    class Solution {
        public boolean isPowerOfThree(int n) {
            return n > 0 && 1162261467 % n == 0;
        }
    }
    
  • // OJ: https://leetcode.com/problems/power-of-three/
    // Time: O(log3(N))
    // Space: O(1)
    class Solution {
    public:
        bool isPowerOfThree(int n) {
            if (n < 1) return false;
            while (n % 3 == 0) n /= 3;
            return n == 1;
        }
    };
    
  • class Solution:
        def isPowerOfThree(self, n: int) -> bool:
            return n > 0 and 1162261467 % n == 0
    
    ############
    
    class Solution(object):
      def isPowerOfThree(self, n):
        """
        :type n: int
        :rtype: bool
        """
        if n > 0:
          return (1162261467 % n) == 0
        else:
          return False
    
    
  • func isPowerOfThree(n int) bool {
    	return n > 0 && 1162261467%n == 0
    }
    
  • function isPowerOfThree(n: number): boolean {
        return n > 0 && 1162261467 % n == 0;
    }
    
    
  • /**
     * @param {number} n
     * @return {boolean}
     */
    var isPowerOfThree = function (n) {
        return n > 0 && 1162261467 % n == 0;
    };
    
    

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