Counting Bits

Clarify the problem:

• The problem requires counting the number of 1 bits in each integer from 0 to a given number.
• We need to implement a function that takes an integer as input and returns an array containing the count of 1 bits for each number from 0 to the input number.

Analyze the problem:

• Input: An integer num.
• Output: An array of integers representing the count of 1 bits for each number from 0 to num.
• Constraints:
  • 0 <= num <= 10^5

Design an algorithm:

• We can solve this problem using a dynamic programming approach.
• Initialize a result array of size num+1 with all elements set to 0.
• Iterate from 1 to num:
  • For each number i, calculate the count of 1 bits by counting the bits in i/2 and adding the least significant bit (i%2).
  • Set the result[i] to the calculated count.
• Return the result array.

Explain your approach:

• We will implement a function called countBits that takes an integer num as input.
• We will initialize a result array of size num+1 with all elements set to 0.
• Then, we will iterate from 1 to num and calculate the count of 1 bits for each number using a dynamic programming approach.
• Finally, we will return the result array.

Write clean and readable code:

python

def countBits(num):
    result = [0] * (num + 1)
    for i in range(1, num + 1):
        result[i] = result[i // 2] + (i % 2)
    return result

Test your code:

python

# Test case 1
# num = 5
# The count of 1 bits for each number from 0 to 5:
# [0, 1, 1, 2, 1, 2]
assert countBits(5) == [0, 1, 1, 2, 1, 2]

# Test case 2
# num = 10
# The count of 1 bits for each number from 0 to 10:
# [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2]
assert countBits(10) == [0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2]

# Test case 3
# num = 0
# The count of 1 bits for each number from 0 to 0:
# [0]
assert countBits(0) == [0]

Optimize if necessary:

• The provided solution has a time complexity of O(num) since we iterate from 1 to num.
• The space complexity is O(num) since we store the result array of size num+1.

Handle error cases:

• The given code assumes that the input num is a non-negative integer. If num is negative, the behavior is undefined.

Discuss complexity analysis:

• The time complexity of the solution is O(num) since we iterate from 1 to num.
• The space complexity is O(num) since we store the result array of size num+1.
• The best-case, worst-case, and average-case scenarios have the same time and space complexity. There are no significant trade-offs in this solution.