0873EasyBit Manipulation

Flipping Every Bit of a Reading

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Trains the technique from

LeetCode 1009Complement of Base 10 Integer

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 meter reading n is held in binary with no leading zeros, so the reading 5 is held as 101.

Flipping the reading turns every 0 into a 1 and every 1 into a 0, across exactly the bits the reading is held in. Flipping 101 gives 010, which is the reading 2.

Return the reading that flipping produces. The reading 0 is held as the single bit 0, so flipping it gives 1.

Examples

Example 1

Input: n = 2
Output: 1

The reading 2 is held as 10. Flipping every bit gives 01, which is the reading 1.

Example 2

Input: n = 8
Output: 7

The reading 8 is held as 1000, and flipping every bit gives 0111.

Example 3

Input: n = 0
Output: 1

The reading 0 is held as the single bit 0, and flipping that bit gives 1.

Constraints

0 <= n <= 999999999

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 flip_reading(n: int) -> int:

Java

public int flipReading(int n)

The intended approach

Time O(log n): the mask doubles each step.Space O(1).

Flipping every bit of a reading held in `k` bits is the same as subtracting it from the number whose `k` bits are all ones, because each bit position contributes to exactly one of the two. So find the smallest all-ones number that reaches at least as far as the reading by doubling and adding one until it does, then subtract. Starting that mask at one rather than zero handles the reading zero, which is held in a single bit.

Pythonentry point: flip_reading

def flip_reading(n):
    mask = 1
    while mask < n:
        mask = mask * 2 + 1
    return mask - n

The obvious version

Write the reading out as binary text, swap each character for the other one, and read the result back as a number. Works on the digits directly rather than with a mask.

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_flip_reading(n):
    text = ""
    left = n
    if left == 0:
        text = "0"
    while left > 0:
        text = ("1" if left % 2 else "0") + text
        left //= 2
    out = 0
    for ch in text:
        out = out * 2 + (0 if ch == "1" else 1)
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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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