1121EasyBit Manipulation

Throws Between Two Switch Banks

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Trains the technique from

LeetCode 2220Minimum Bit Flips to Convert Number

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.

Two banks of switches are described by the whole numbers start and goal. A switch is on wherever the number carries a 1 bit, switches are numbered from the lowest bit upwards, and any place beyond a number's own bits counts as off.

One throw flips a single switch. Return the fewest throws that turn the start bank into the goal bank.

Examples

Example 1

Input: start = 15, goal = 240
Output: 8

The lowest four switches are on in the first bank and off in the second, and the next four are the other way round, so all eight have to be thrown.

Example 2

Input: start = 1023, goal = 1024
Output: 11

The first bank has its lowest ten switches on while the second has only the eleventh, so ten switches go off and one comes on.

Example 3

Input: start = 1, goal = 1
Output: 0

The two banks already match, so nothing is thrown.

Constraints

0 <= start <= 10^9
0 <= goal <= 10^9

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 min_bit_flips(start: int, goal: int) -> int:

Java

public int minBitFlips(int start, int goal)

The intended approach

Time O(b) over the bit places of the two numbers.Space O(1).

Each switch stands alone: a throw changes one switch and nothing else, so a switch already matching needs no throw and a switch differing needs exactly one. The answer is therefore just how many places the two banks disagree in, with no ordering or interaction to consider. Exclusive-or is precisely the operation that marks the disagreeing places, leaving a `1` where the two banks differ and a `0` where they match. So combine the two numbers that way and count the `1` bits of the result by testing the lowest bit and shifting down until nothing is left. Places above both numbers are off in both, so they never contribute and the loop can stop as soon as the remainder runs out.

Pythonentry point: min_bit_flips

def min_bit_flips(start, goal):
    gap = start ^ goal
    throws = 0
    while gap:
        throws += gap & 1
        gap >>= 1
    return throws

The obvious version

Read out each switch of both banks in turn by dividing down to the place and taking the remainder against two, then count the places where the two answers differ, walking a fixed forty places which is more than either number can need.

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_min_bit_flips(start, goal):
    throws = 0
    for place in range(40):
        step = 2 ** place
        if (start // step) % 2 != (goal // step) % 2:
            throws += 1
    return throws

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 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

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