Problem Description

Given an integer n, you need to create an array of length n + 1 where each element represents the count of 1s in the binary representation of its index.

For each index i from 0 to n (inclusive), the value at ans[i] should be the number of 1 bits in the binary representation of i.

For example:

• If n = 5, you need to count the 1s in binary representations of 0, 1, 2, 3, 4, 5
• 0 in binary is 0 → has 0 ones
• 1 in binary is 1 → has 1 one
• 2 in binary is 10 → has 1 one
• 3 in binary is 11 → has 2 ones
• 4 in binary is 100 → has 1 one
• 5 in binary is 101 → has 2 ones
• So the output would be [0, 1, 1, 2, 1, 2]

The solution uses Python's built-in bit_count() method which directly counts the number of 1 bits in the binary representation of each number from 0 to n. This is done using a list comprehension that iterates through the range [0, n+1) and applies bit_count() to each value.

Intuition

The most straightforward way to solve this problem is to count the 1 bits for each number from 0 to n. Since we need to find the bit count for every number in sequence, we can iterate through each number and calculate its bit count.

Python provides a built-in method bit_count() that directly counts the number of set bits (1s) in an integer's binary representation. This eliminates the need to manually convert numbers to binary strings or implement bit manipulation logic ourselves.

The key insight is that we don't need to find any pattern or optimization here - we can simply compute what's asked directly. For each number i in the range [0, n], we call i.bit_count() to get the number of 1s in its binary form.

While there are more complex approaches using dynamic programming (like observing that the bit count of i relates to the bit count of i//2 or i & (i-1)), the direct approach using bit_count() is both simple and efficient. It leverages Python's optimized built-in functionality to solve the problem in a single line using list comprehension.

The time complexity remains O(n) as we iterate through all numbers once, and the space complexity is O(n) for storing the result array, which is optimal since we need to return an array of size n + 1.

Solution Approach

The implementation uses a list comprehension to create the result array in a single line. Let's break down the solution:

def countBits(self, n: int) -> List[int]:
    return [i.bit_count() for i in range(n + 1)]

Step-by-step breakdown:

1. Range Generation: range(n + 1) creates a sequence of numbers from 0 to n (inclusive). For example, if n = 5, this generates [0, 1, 2, 3, 4, 5].

2. Bit Counting: For each number i in this range, we call i.bit_count(). This built-in Python method (available from Python 3.10+) efficiently counts the number of 1 bits in the binary representation of the integer.

3. List Comprehension: The syntax [expression for item in iterable] creates a new list by applying the expression to each item. Here, we apply bit_count() to each number in our range.

How bit_count() works internally:

• For i = 0: binary is 0, so bit_count() returns 0
• For i = 1: binary is 1, so bit_count() returns 1
• For i = 2: binary is 10, so bit_count() returns 1
• For i = 3: binary is 11, so bit_count() returns 2
• And so on...

The entire computation happens in a single pass through the range, making this solution both concise and efficient. The method directly returns the constructed list without needing any intermediate variables or complex logic.

Time Complexity: O(n) - we iterate through n + 1 numbers once Space Complexity: O(n) - we create an array of size n + 1 to store the results

Example Walkthrough

Let's walk through the solution with n = 6 to see how the algorithm works step by step.

Step 1: Generate the range We need to process numbers from 0 to 6 (inclusive), so our range is [0, 1, 2, 3, 4, 5, 6].

Step 2: Apply bit_count() to each number

Let's trace through each iteration:

• i = 0:

  • Binary representation: 0
  • Count of 1s: 0
  • Result array so far: [0]

• i = 1:

  • Binary representation: 1
  • Count of 1s: 1
  • Result array so far: [0, 1]

• i = 2:

  • Binary representation: 10
  • Count of 1s: 1
  • Result array so far: [0, 1, 1]

• i = 3:

  • Binary representation: 11
  • Count of 1s: 2
  • Result array so far: [0, 1, 1, 2]

• i = 4:

  • Binary representation: 100
  • Count of 1s: 1
  • Result array so far: [0, 1, 1, 2, 1]

• i = 5:

  • Binary representation: 101
  • Count of 1s: 2
  • Result array so far: [0, 1, 1, 2, 1, 2]

• i = 6:

  • Binary representation: 110
  • Count of 1s: 2
  • Final result: [0, 1, 1, 2, 1, 2, 2]

Step 3: Return the result The list comprehension [i.bit_count() for i in range(7)] produces [0, 1, 1, 2, 1, 2, 2], which is our answer.

The solution elegantly handles all cases in a single expression, automatically building the result array as it processes each number in sequence.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The code iterates through all integers from 0 to n (inclusive), which gives us n + 1 iterations. For each integer i, the bit_count() method is called. In Python 3.10+, bit_count() is implemented efficiently with a time complexity of O(1) using bit manipulation techniques (population count). Therefore, the overall time complexity is O(n + 1) * O(1) = O(n).

Space Complexity: O(n)

The space complexity consists of:

• Output space: The returned list contains n + 1 elements, requiring O(n) space.
• Auxiliary space: The list comprehension doesn't use any additional data structures beyond the output list itself, so auxiliary space is O(1).

Since we typically include the output space in the total space complexity for this type of problem, the overall space complexity is O(n).

Common Pitfalls

1. Python Version Compatibility Issue

The bit_count() method is only available in Python 3.10 and later versions. If you're using an older Python version or submitting to a platform that doesn't support Python 3.10+, this solution will fail with an AttributeError.

Solution: Use alternative approaches that work with older Python versions:

# Option 1: Using bin() and count()
def countBits(self, n: int) -> List[int]:
    return [bin(i).count('1') for i in range(n + 1)]

# Option 2: Using bit manipulation (Brian Kernighan's algorithm)
def countBits(self, n: int) -> List[int]:
    def count_ones(num):
        count = 0
        while num:
            num &= num - 1  # Removes the rightmost 1-bit
            count += 1
        return count
    return [count_ones(i) for i in range(n + 1)]

# Option 3: Dynamic Programming approach
def countBits(self, n: int) -> List[int]:
    ans = [0] * (n + 1)
    for i in range(1, n + 1):
        ans[i] = ans[i >> 1] + (i & 1)
    return ans

2. Off-by-One Error with Range

A common mistake is using range(n) instead of range(n + 1), which would miss counting the bits for the number n itself.

Incorrect:

return [i.bit_count() for i in range(n)]  # Missing the last element

Correct:

return [i.bit_count() for i in range(n + 1)]  # Includes n

3. Memory Optimization Misconception

While the one-liner solution is elegant, developers might think they need to optimize memory by computing values on-the-fly. However, since the problem requires returning the entire array, there's no memory advantage to computing values individually versus using list comprehension.

4. Performance Assumptions

Some might assume that using built-in methods like bit_count() is always faster than bit manipulation techniques. While bit_count() is highly optimized, for competitive programming or interview settings, demonstrating knowledge of bit manipulation patterns (like using dynamic programming with previously computed values) can be valuable:

# DP solution that leverages previously computed values
def countBits(self, n: int) -> List[int]:
    ans = [0] * (n + 1)
    for i in range(1, n + 1):
        # i >> 1 gives i//2, and i & 1 gives the last bit
        ans[i] = ans[i >> 1] + (i & 1)
    return ans

This DP approach has the same O(n) time complexity but demonstrates understanding of the pattern that the bit count of i equals the bit count of i//2 plus the last bit of i.