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231_powerOfTwo.md

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231. Power of Two 🌟

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

An integer n is a power of two, if there exists an integer x such that n == 2x.

Code

class Solution {
public:
    bool isPowerOfTwo(int n) {
        if(n<=0)return false;
        return ((n&(n-1))==0);
    }
};

Method 1: Iterative

check if n can be divided by 2. If yes, divide n by 2 and check it repeatedly.

if (n <= 0) return false;
while (n%2 == 0) n/=2;
return n == 1;

Time complexity = O(log n)

Method 2: Recursive

return n > 0 && (n == 1 || (n%2 == 0 && isPowerOfTwo(n/2)));

Time complexity = O(log n)

Method 3: Bit operation

If n is the power of two:

n = 2 ^ 0 = 1 = 0b0000...00000001, and (n - 1) = 0 = 0b0000...0000.
n = 2 ^ 1 = 2 = 0b0000...00000010, and (n - 1) = 1 = 0b0000...0001.
n = 2 ^ 2 = 4 = 0b0000...00000100, and (n - 1) = 3 = 0b0000...0011.
n = 2 ^ 3 = 8 = 0b0000...00001000, and (n - 1) = 7 = 0b0000...0111.
we have n & (n-1) == 0b0000...0000 == 0
Otherwise, n & (n-1) != 0.

For example, n =14 = 0b0000...1110, and (n - 1) = 13 = 0b0000...1101.

return n > 0 && ((n & (n-1)) == 0);

Time complexity = O(1)

Method 4: Math derivation

Because the range of an integer = -2147483648 (-2^31) ~ 2147483647 (2^31-1), the max possible power of two = 2^30 = 1073741824.

(1) If n is the power of two, let n = 2^k, where k is an integer.

We have 2^30 = (2^k) * 2^(30-k), which means (2^30 % 2^k) == 0.

(2) If n is not the power of two, let n = j*(2^k), where k is an integer and j is an odd number.

We have (2^30 % j*(2^k)) == (2^(30-k) % j) != 0.

return n > 0 && (1073741824 % n == 0);

Time complexity = O(1)

Method 5: Bit count

Very intuitive. If n is the power of 2, the bit count of n is 1. Note that 0b1000...000 is -2147483648, which is not the power of two, but the bit count is 1.

return n > 0 && Integer.bitCount(n) == 1;

Time complexity = O(1) The time complexity of bitCount() can be done by a fixed number of operations. More info in https://stackoverflow.com/questions/109023.

Method 6: Look-up table

There are only 31 numbers in total for an 32-bit integer.

return new HashSet<>(Arrays.asList(1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 
1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 
2097152, 4194304, 8388608,16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824)).contains(n);

time complexity = O(1)

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