What is the brute force approach and complexity for Counting Bits?

For each i, shift it right and add up its bits with time complexity O(N log N) and space complexity O(1).

What is the optimal approach and complexity for Counting Bits?

ans[i] = ans[i >> 1] + (i & 1), filled from 1 up to n with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for Counting Bits?

Look for key signals in the prompt: "number of 1 bits for each i from 0 to n" indicates Many answers, one per number. Reuse earlier answers instead of counting from scratch.; "do not use built-in bit counting" indicates The shift rule itself has to be found: `i >> 1` plus the last bit.; "linear time, single pass" indicates Each number is solved from one smaller number, in increasing order..

Counting Bits

Easy·LeetCode #338 Pro OnlySlight Variation

LeetCode

Problem Statement

Given n, build an array of n + 1 numbers. The number at index i is how many 1 bits i has in binary, for every i from 0 to n. Do not use built-in bit-count functions.

Follow-up: it is easy to get O(N log N) time. Can you do it in O(N) time, in a single pass?

Input

An integer n.

Output

An array ans of length n + 1, where ans[i] is the number of 1 bits in i.

Example 1

Input: n = 2

Output: [0, 1, 1]

0 has no 1 bits. 1 = 1 has one. 2 = 10 has one.

Example 2

Input: n = 5

Output: [0, 1, 1, 2, 1, 2]

3 = 11 has two 1 bits, 4 = 100 has one, and 5 = 101 has two.

Constraints

• 0 ≤ n ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“number of 1 bits for each i from 0 to n”

→ Many answers, one per number. Reuse earlier answers instead of counting from scratch.

“do not use built-in bit counting”

→ The shift rule itself has to be found: `i >> 1` plus the last bit.

“linear time, single pass”

→ Each number is solved from one smaller number, in increasing order.

Approach Complexity Ladder

Brute Force — count each number's bitsO(N log N)O(1)

For each i, shift it right and add up its bits

Optimal — reuse the answer for i >> 1O(N)O(N)

ans[i] = ans[i >> 1] + (i & 1), filled from 1 up to n

Brute Force Intuition

Count the 1 bits of each i from 0 to n on its own. Each count shifts i right one bit at a time and adds the lowest bit.

Recognition rule for this sub-pattern:

n & (n−1) removes the lowest set bit; shifts read bits one at a time.

Remember this · Bit Manipulation

Trigger

"Every element appears twice except…", power of two, counting set bits, subsets as bitmasks, O(1) space tricks.

Think

What happens to this problem if I look at each bit separately?

Key question

Which bit operation (XOR, AND, OR, shift) cancels or isolates what I need?

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