Counting Bits

The problem requires us to count the number of 1’s in the binary representation of numbers from 0 to n and return the counts in an array.

We can use dynamic programming (DP) approach to solve this problem in (O(n)) time. Here’s a step-by-step approach:

  1. Initialize an Array: Initialize an array ans of length n + 1 to store the counts of 1’s.
  2. Set Base Cases: ans[0] will be 0 as there are no 1’s in the binary representation of 0.
  3. Iterate Through Numbers: Iterate from 1 to n, and for each i, you can calculate ans[i] as ans[i & (i - 1)] + 1. Here, i & (i - 1) removes the rightmost set bit of i, and as we have already calculated the number of 1’s for i & (i - 1), we just need to add 1 to that value.

Here’s the code:

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class Solution:
    def countBits(self, n: int) -> List[int]:
        ans = [0] * (n + 1)
        for i in range(1, n + 1):
            ans[i] = ans[i & (i - 1)] + 1
        return ans

Key Takeaways

  • This code uses the bitwise operation to calculate the number of 1’s in the binary representation efficiently.
  • The time complexity of this code is (O(n)), and the space complexity is also (O(n)).
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def count_ones(n)
    output = 0
    while n != 0
       output += 1
       n = n & (n-1)
    end
    output
end

0 <= i < 2

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# @param {Integer} num
# @return {Integer[]}
def count_bits(num)
    output = []
    for i in (0..num)
       output << count_ones(i) 
    end
    output
end

Identifying Problem Isomorphism

“Counting Bits” can be approximately mapped to “Number of 1 Bits”.

Reasoning:

Both problems involve analyzing the binary representation of numbers. They are related to bit manipulation.

In “Counting Bits”, the task is to count the number of 1’s in the binary representation of all numbers in the range from 0 to a given number.

In “Number of 1 Bits”, the task is to write a function that takes an unsigned integer and returns the number of ‘1’ bits it has.

The difference is that “Number of 1 Bits” focuses on a single number, while “Counting Bits” involves a range of numbers. They share a fundamental concept of counting the number of 1’s in binary representations, making these problems approximately isomorphic.

“Number of 1 Bits” is simpler as it requires counting bits for a single number. “Counting Bits” requires counting bits for a range of numbers, adding a layer of complexity.

“Counting Bits” is about bit manipulation and dynamic programming. Here are problems to prepare for this one:

  1. 191. Number of 1 Bits: This problem involves counting the number of 1 bits, which is an essential part of the “Counting Bits” problem.

  2. 231. Power of Two: Understanding how power of two numbers are represented in binary will help in “Counting Bits”.

  3. 136. Single Number: This problem practices bitwise XOR operation, an important concept for bit manipulation.

  4. 190. Reverse Bits: This problem provides practice for bit manipulation.

  5. 371. Sum of Two Integers: This problem requires understanding of bit operations to simulate addition without using ‘+’ or ‘-’.

  6. 389. Find the Difference: More practice on bit manipulation operations.

  7. 268. Missing Number: This problem also uses bit manipulation to find the missing number.

  8. 137. Single Number II: A more complex problem related to bit manipulation which will improve understanding of bitwise operations.

  9. 70. Climbing Stairs: A simpler problem to understand the concept of dynamic programming.

These cover bitwise manipulation and dynamic programming strategies which are key for the “Counting Bits” problem.

Clarification Questions

What are the clarification questions we can ask about this problem?

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def countBits(self, num: int) -> List[int]:
    counter = [0]

    for i in range(1, num+1):
        counter.append(counter[i >> 1] + i % 2)
    return counter

Problem Classification

This problem can be classified into several categories:

  1. Computer Science and Information Theory: The problem involves dealing with binary representations of numbers, which is a foundational concept in computer science and information theory.

  2. Counting Problem: The problem is about counting the number of 1’s in the binary representation of a number, making it a counting problem.

  3. Array Processing: As you are required to return an array with counts for every number up to n, it involves array creation and manipulation.

  4. Dynamic Programming: The problem can be solved more efficiently using dynamic programming by using previously computed values (number of 1’s in the binary representation of i-1) to determine the current value.

Language Agnostic Coding Drills

  1. Understanding Binary Operations: Binary operations are at the core of this problem. The two operations used here are bitwise right shift (») and modulus (%). The right shift operation essentially halves the number, moving all digits one place to the right. The modulus operation here is used to check if a number is odd or even (odd numbers have a 1 in the least significant bit).

    Drill: Write a code to perform right shift operation and modulus operation on a number.

  2. List Initialisation and Indexing: The algorithm begins by initialising a list with a single element, 0. The elements of this list represent the count of 1’s in the binary representation of the index of the element.

    Drill: Create a list and practise basic operations like initialisation, appending, and accessing elements via indexing.

  3. Loop Control Structures: The problem uses a loop to iterate over each number from 1 up to n. Understanding how to control and use loops is key here.

    Drill: Write a loop that iterates from 1 to n, printing each number.

  4. List Manipulation and Appending: As each iteration is made, a new value is computed and appended to the list. This drill focuses on manipulating lists within a loop.

    Drill: Within a loop, perform calculations and append the results to a list.

  5. Dynamic Programming: This problem can be solved with a dynamic programming approach. It reuses previously calculated results to calculate the current result.

    Drill: Write a simple dynamic programming algorithm, such as computing Fibonacci numbers, where each number is the sum of the two preceding ones.

In the final solution, the goal is to fill up the list with the number of 1’s in the binary representation of each number from 0 to n. This is done by iteratively calculating the number of 1’s in each number, where the number of 1’s in a number is the number of 1’s in the number obtained by right shifting the current number once (which is equivalent to halving the number, or discarding the least significant bit in binary) plus the least significant bit in the current number. This approach works because each binary number is just a previous binary number shifted to the left (multiplied by 2) with a new bit added at the end.

Targeted Drills in Python

  1. Understanding Binary Operations:

    Here is a simple drill where we will perform right shift and modulus operations.

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    def binary_operations(num):
    print("Original number: ", num)
    print("Number after right shift: ", num >> 1)
    print("Number modulus 2: ", num % 2)

binary_operations(10)

  2. List Initialisation and Indexing:

    In this drill, we will create a list, append some values and access them.

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    def list_operations():
    my_list = [0]
    print("Initial list: ", my_list)

    # Append some values
    for i in range(1, 10):
        my_list.append(i)
    print("List after appending values: ", my_list)

    # Access elements via indexing
    print("Element at index 5: ", my_list[5])

list_operations()

  3. Loop Control Structures:

    This drill is a simple loop printing each number from 1 to n.

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    def loop_structure(n):
    for i in range(1, n+1):
        print(i)

loop_structure(10)

  4. List Manipulation and Appending:

    In this drill, we perform calculations within a loop and append the results to a list.

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    def list_manipulation(n):
    result_list = []
    for i in range(1, n+1):
        result_list.append(i**2)
    print(result_list)

list_manipulation(10)

  5. Dynamic Programming:

    This drill is a simple dynamic programming algorithm that calculates Fibonacci numbers.

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    def fibonacci(n):
    fib_values = [0, 1] + [0] * (n - 1)
    for i in range(2, n + 1):
        fib_values[i] = fib_values[i - 1] + fib_values[i - 2]
    return fib_values[n]

print(fibonacci(10))

Each of these drills focuses on a particular aspect of the final solution. The final solution will integrate all these drills into one cohesive solution by creating a list of length n+1, then iterating over each number from 1 to n, and for each iteration, calculating the number of 1’s in the binary representation of the current number and appending that to the list. The solution employs the concepts of binary operations, list manipulation, and dynamic programming.