Power of Four | LeetCode 342 | C++, Java, Python

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Комментарии

Amazing, I've learned so much from you. Thank you so much for all of your content.

crisag.

Improved audio quality, presentation and explanation method. This man is on a roll.
Keep Making !

acxd

Thanks for this video. Great explanation.

akshatmathur

Very nicely explained!! Thank you so much.

singhrav

Wake up at morning...watch knowledge center daily leetcode challenge videos..do extra questions on leetcode...go to sleep

raviashwin

Thank you. None has reached your level at explaining leetcode so far.

mohammadyahya

very helpful! thank you for the great explanation

jacksonwahl

use log function
class Solution:
def isPowerOfFour(self, num):
import math
if num>0:
a=math.log(num)/math.log(4)
b=a-int(a)
else:
return False
if b==0:
return True

mohitarora

Please add #bitmanipulation so that searching all bit manipulation videos will be easy

Starfalll

My Solution
Time- O(1), Space- O(1)
bool isPowerOfFour(int num) {
if(num <= 0)
return false;
int val1 = floor(log2(num)), val2 = ceil(log2(num));
if(val1 != val2)
return false;
return !(val1&1);
}

rishisharma

Bro where should I focus on codeforces or codechef

manu-singh

can we take log (base 4) in c++....if yes then how

shivanktripathi

class Solution {
public:
bool isPowerOfFour(int num1) {
if(num1 == 1)
return true;
long long int num = num1;
if((num & num - 1) == 0 && (num % 10 == 4 || num%10 == 6 ))
return true;
return false;
}
};
Explanation:
each power of 4 has either 4 or 6 one's digit. And power(4) is a subset of power(2) and we can solve power(2) as (n & n-1) == 0 and check whether it has units digit 4 and 6

deepakkumarsahu

what is "x" in dont mind im yet a beginner hoping for clarification of this doubt

pillabindu

try:
return int(math.log(num, 4)) == (math.log(num, 4))
except ValueError:
pass

midhileshmomidi

return (num>0 && (num & (num-1)) == 0 && (int)log2(num) % 2 == 0);

JardaniJovonovich

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