Skip to main content

Power of Four

Power of Four 




Topic : Recursion

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

An integer n is a power of four, if there exists an integer x such that n == 4x.



    bool isPowerOfFour(int n) {
        if(n<=0) return false;
        if(n==1 || n==4) return true;
        if(n%4 !=0) return false;
        return isPowerOfFour(n/4);
    }


public boolean isPowerOfFour(int n) {
    return n > 1 && n %4 == 0 ? isPowerOfFour(n/4) : n == 1;
}


class Solution:
    def isPowerOfFour(self, n: int) -> bool:
        if n == 1: 
            return True 
        elif n == 0: 
            return False
        else: 
            return n%4 == 0 and self.isPowerOfFour(n//4)


Wondering :

class Solution: def isPowerOfFour(self, n: int) -> bool: return n > 0 and log2(n) % 2 == 0






























Comments

Popular posts from this blog

Leetcode 371. Sum of Two Integers. C++ / Java

371 .  Sum of Two Integers   Given two integers  a  and  b , return  the sum of the two integers without using the operators   +   and   - .   Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5   Constraints: -1000 <= a, b <= 1000 Solution :  C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java :  class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Leetcode 347. Top K Frequent Elements. Python (Bubble Sort)

Top K Frequent Elements Given an integer array  nums  and an integer  k , return  the   k   most frequent elements . You may return the answer in  any order .   Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1]   Constraints: 1 <= nums.length <= 10 5 -10 4  <= nums[i] <= 10 4 k  is in the range  [1, the number of unique elements in the array] . It is  guaranteed  that the answer is  unique .   Follow up:  Your algorithm's time complexity must be better than  O(n log n) , where n is the array's size. Solution : class Solution:     def topKFrequent(self, n: List[int], k: int) -> List[int]:                  # [1,1,1,2,2,3] &  k = 2                  f = [[] for i in range(len(n) + 1)]         # f = [...

Container with Most Water - Leetcode 11 - Python

  You are given an integer array   height   of length   n . There are   n   vertical lines drawn such that the two endpoints of the   i th   line are   (i, 0)   and   (i, height[i]) . Find two lines that together with the x-axis form a container, such that the container contains the most water. Return  the maximum amount of water a container can store . Notice  that you may not slant the container.   Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49. class Solution:     def maxArea(self, height: List[int]) -> int:         # ans = 0         # for l in range(len(height)):         #     for r in range(l+1, len(height)):         #      ...