0571EasyDynamic ProgrammingBit Manipulation

Lit Lamps Per Ticket Id

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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 queue display shows a ticket id in binary on a row of lamps: a lamp is lit for each 1 in the binary form of the id, and dark for each 0. Ids run from 0 up to n inclusive.

Report the lit lamp count for each id in turn, starting at id 0 and finishing at id n, so the reported list holds one more entry than n. Id 0 lights nothing.

Aim for a running time linear in n, without a fresh scan of the bits of every id.

Examples

Example 1

Input: n = 9
Output: [0, 1, 1, 2, 1, 2, 2, 3, 1, 2]

Ids 0 to 9 in binary are 0, 1, 10, 11, 100, 101, 110, 111, 1000 and 1001, so the lit lamp counts run 0, 1, 1, 2, 1, 2, 2, 3, 1, 2.

Example 2

Input: n = 13
Output: [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3]

The counts for ids 0 to 9 are unchanged, and ids 10 to 13 are 1010, 1011, 1100 and 1101, lighting 2, 3, 2 and 3 lamps.

Example 3

Input: n = 0
Output: [0]

Only id 0 exists and it lights no lamp, so the answer holds a single zero.

Constraints

0 <= n <= 10^5
The returned array has n + 1 entries, and every entry is at most 17.

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 count_bits(n: int) -> list[int]:

Java

public int[] countBits(int n)

The intended approach

Time O(n).Space O(n) for the returned table, and nothing else.

Fill a table from index 1 upward. Shifting an id right by one bit gives a strictly smaller id whose entry is already filled in, and the bit that fell off is exactly the lamp that separates the two, so each entry is one earlier entry plus the id's lowest bit.

Pythonentry point: count_bits

def count_bits(n):
    # O(n) time, O(n) space for the returned table: dropping the last bit of i
    # leaves a smaller index whose answer is already known, so each entry costs a
    # shift, a lookup and one addition.
    table = [0] * (n + 1)
    for i in range(1, n + 1):
        table[i] = table[i >> 1] + (i & 1)
    return table

The obvious version

For every id from 0 to n, halve it repeatedly and add up the remainders, which counts its lit lamps from scratch.

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_count_bits(n):
    out = []
    for i in range(n + 1):
        v = i
        lit = 0
        while v > 0:
            lit += v % 2
            v //= 2
        out.append(lit)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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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