Counting Bits Leetcode Problem 338 [Python Solution]

Welcome to another coding adventure! In this post tutorial, we’ll be Counting Bits LeetCode problem, which is categorized as an easy bit manipulation problem.

That didn’t sound very right, but had to get it in there.

The problem statement goes like this: given an integer n, we need to return an array, ans, of length n + 1, where each element ans[i] (0 <= i <= n) represents the number of 1s in the binary representation of i.

To put it simply, we want to count how many ones there are in the binary representation of each integer from 0 to n.

Problem Overview

Let’s break down the problem using an example:

Example 1:

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

Here’s the explanation:

• 0 in binary is 0, which has 0 ones.
• 1 in binary is 1, which has 1 one.
• 2 in binary is 10, which also has 1 one.

Now that we understand the problem, let’s dive deeper into the solution.

Understanding Constraints

Before we proceed with the solution, let’s take a moment to understand the problem’s constraints:

0 <= n <= 105: This tells us that n can be any non-negative integer up to 105.

Edge Cases a Valid Solution Must Consider

When n is 0, the output should be a list with a single element, 0, as there are no ones in the binary representation of 0.

Now, let’s explore the solutions for this problem.

Counting Bits LeetCode Problem Solution

1. Bruteforce Approach

One way to solve this problem is by using a brute-force approach.

We iterate through all the numbers from 0 to n, convert each number to binary, and count the number of ones in the binary representation.

Here’s a Python code snippet to demonstrate this approach:

def countBitsBruteforce(n: int):
    ans = []
    for i in range(n + 1):
        # Count the number of ones in the binary representation of i
        ones_count = bin(i).count('1')
        ans.append(ones_count)
    return ans

While this approach works, it’s not the most efficient solution.

The time complexity of this approach is O(n * log(n)), which is not optimal for large values of n.

2. An Efficient Approach with Dynamic Programming

To optimize our solution, we can use dynamic programming.

The key idea here is to notice a pattern in how we can compute the number of ones for each integer based on the results we’ve computed for previous integers.

Let’s illustrate this with an example:

Suppose we’ve computed the number of ones for integers from 0 to 7:

• 0: 0 ones
• 1: 1 one
• 2: 1 one
• 3: 2 ones
• 4: 1 one
• 5: 2 ones
• 6: 2 ones
• 7: 3 ones

We observe that for each new integer, the number of ones can be determined as follows:

• For the least significant bit, we can simply count the ones in the binary representation. This part is straightforward.

• For the more significant bits, we can utilize the results we’ve already calculated.

Let’s see how:

For example, when we calculate the number of ones for integer 6, we can see that it has 2 ones in its binary representation.

But we also know that integer 4 has 1 one, and if we add 2 to that (since we’re dealing with the next power of 2), we get the number of ones in integer 6. This leads us to an efficient dynamic programming solution.

Here’s the Python code for it:

def countBits(n: int):
    # Initialize an array to store the results
    dp = [0] * (n + 1)

    # Offset is used to track the most significant bit
    offset = 1

    for i in range(1, n + 1):
        # If offset * 2 == i, update the offset
        if offset * 2 == i:
            offset = i
        # Calculate the number of ones based on the offset
        dp[i] = 1 + dp[i - offset]

    return dp

This efficient dynamic programming approach has a time complexity of O(n) and a space complexity of O(n), making it a much better choice for larger values of n.

Reasoning Behind Our Approach

We’ve optimized our solution by noticing a pattern in how the number of ones in the binary representation of an integer can be determined based on the results we’ve calculated for previous integers.

This dynamic programming approach allows us to compute the answer in linear time, making it efficient for large inputs.

Time and Space Complexity

• Time Complexity: O(n)
• Space Complexity: O(n)

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Conclusion

In this blog post, we’ve explored the Counting Bits LeetCode problem, which asks us to count the number of ones in the binary representation of integers from 0 to n.

We’ve discussed two approaches: a brute-force approach and an efficient dynamic programming approach.

The efficient dynamic programming solution is recommended for this problem as it has a time complexity of O(n) and a space complexity of O(n), making it suitable for larger input values.

By understanding the problem, the constraints, and the edge cases, we can efficiently solve a wide range of coding problems.

Now that you have a solid grasp of how to solve the Counting Bits problem, I encourage you to practice your skills and explore more LeetCode challenges.

Feel free to comment, ask questions, make suggestions, and share this content.

Happy coding!

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