0388EasyDivide and ConquerBit Manipulation

Energised Relay Count

Tracked in this browser only

Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A test rig drives a bank of relays from one control register. The register holds the positive integer state, and relay k is energised exactly when the bit of weight 2^k is set in state.

Return how many relays are energised.

Examples

Example 1

Input: state = 26
Output: 3

26 is 11010 in binary, and three of those bits are set.

Example 2

Input: state = 512
Output: 1

512 is 2^9, so only the relay of weight 512 is energised.

Example 3

Input: state = 4095
Output: 12

4095 is 111111111111 in binary, twelve set bits in a row.

Example 4

Input: state = 1073741824
Output: 1

1073741824 is 2^30, so a single relay near the top of the bank is energised.

Constraints

1 <= state <= 2^31 - 1

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def hamming_weight(state: int) -> int:

Java

public int hammingWeight(int state)

The intended approach

Time O(k) for k energised relays, at most 31 for the stated rangeSpace O(1)

`state & (state - 1)` clears the lowest set bit of `state` and leaves every other bit alone, because subtracting one flips that lowest set bit to 0 and turns the zeros below it into ones. Applying it repeatedly until the register reaches zero therefore runs exactly once per energised relay, so counting the iterations gives the answer.

Pythonentry point: hamming_weight

def hamming_weight(state):
    # Time O(k) where k is the number of energised relays, so at most 31 iterations
    # for the stated range, i.e. O(1) here.
    # Space O(1).
    count = 0
    while state:
        state &= state - 1
        count += 1
    return count

The obvious version

Divide the register by two until it reaches zero, adding one whenever the remainder is odd. Visits every bit position rather than only the set ones.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_hamming_weight(state):
    count = 0
    value = state
    while value > 0:
        if value % 2 == 1:
            count += 1
        value //= 2
    return count

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in