Check if all bits of a number are set

Given a number n, check whether every bit in the binary representation of the given number is set or not, if yes return true else return false.

Examples :

Input: n = 7
Output: true
Explanation: Binary for 7 is 111, all the bits are set so output is true.

Input: n = 8
Output: false
Explanation: Binary for 8 is 1000, all the bits are not set so output is false.

[Naive Approach] Bits by Bit Traversal - O(log n) Time and O(1) Space

The intuition is to check every bit one by one starting from the right. If any bit is 0 (not set), return false immediately. If you get through the whole number and every bit was 1, return true.

Consider the following dry run: n = 7, binary representation of 7 : 111

For n = 7 (111), (n & 1) = 1 ---> OK, right shift by 1 ---> n = 3 (011)
For n = 3(011), (n & 1) = 1 ---> OK, right shift by 1 ---> n = 1 (001)
For n = 1(001), (n & 1) = 1 ---> OK, right shift by 1 ---> n = 0(000)

Loop ends because n = 0, and no 0 bit encountered so returns true.

Final answer : true

true

[Alternate Approach] Check for (2^k − 1) - O(log n) Time and O(1) Space 

The idea is that a number whose binary representation has all bits set (like 7 → 111, 15 → 1111) is always of the form 2^k - 1, where k is the number of set bits. So, we first count the number of set bits in the number. Let this count be k. Then we construct the number (2^k - 1), which is a number having exactly k bits set (like 111...k times). If the original number n is equal to (2^k - 1), it means all its bits are set. Otherwise, it is not.

Consider the following dry run : n = 7 ---> (111)

Step 1: n = 7 ---> n == 0? NO ---> continue
Step 2: Count set bits in the number n, (since 111 has 3 set bits) so count = 3
Step 3: (1 << count) = (1 << 3) = 8 ---> (1000)
Step 4: (1 << count) - 1 = 8 - 1 = 7 ---> (111)
Step 5: Compare n == 7 ---> TRUE
Step 6: return true

Final answer : true

true

[Optimal Approach] Using Check for Power of Two - O(1) Time and O(1) Space

The idea is that if a number n has all bits set in its binary representation (like 7 → 111), then adding 1 to it will produce a power of two (like 8 → 1000).

This happens because adding 1 to a sequence of all 1s causes a carry to propagate through all bits, turning them into 0s and adding a new leading 1. So, we compute x = n + 1 and then check whether x is a power of 2.

To check if a number is a power of 2, we use the condition (x & (x - 1)) == 0, which works because powers of 2 have exactly one bit set in binary.

Consider the following dry run : n = 7 --> (111)

Step 1: n = 7 ---> n <= 0? NO ---> continue
Step 2: x = n + 1 = 7 + 1 = 8 ---> (1000)
Step 3: x - 1 = 8 - 1 = 7 ---> (0111)
Step 4: (x & (x - 1)) = 1000 & 0111 = 0000 ---> equals 0
Step 5: return true

Final answer : true

true