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Code Bit Logic

# Count total bits in a number

The problem we are addressing involves counting the total number of bits in a given integer. In binary representation, an integer is composed of a series of bits, which can be either 0 or 1. The task is to develop a method to count how many bits are present in the binary representation of a number.

## Problem Statement and Description

Given an integer `n`, the objective is to determine the total number of bits in its binary representation. In other words, we want to count how many digits are required to represent the number in binary form. This task can be accomplished by utilizing logarithmic calculations.

## Example

Consider the integer 7. In binary, 7 is represented as 111. The binary representation contains 3 bits. Thus, the expected output for this example should be "Bits: 3".

Another example is 16. In binary, 16 is represented as 10000. The binary representation contains 5 bits. Therefore, the expected output for this example should be "Bits: 5".

``````Number : 34
Output : 6
[34 : 100010]
|<------>|  = 6
Number : 256
Output : 9
[255 : 100000000]
|<--------->|  = 9
``````

## Idea to Solve the Problem

The fundamental idea to solve this problem is to use logarithms to calculate the number of bits. Specifically, we can utilize the base-2 logarithm (binary logarithm) to determine the minimum number of bits required to represent a given number.

## Pseudocode

Here's the pseudocode for the algorithm:

``````function countBits(n):
if n < 1:
return
result = floor(log2(n) + 1)
print "Number:", n
print "Bits:", result``````

## Algorithm Explanation

1. Start by checking if the given number `n` is less than 1. If it is, return since the concept of counting bits doesn't apply to non-positive numbers.
2. If `n` is greater than or equal to 1, calculate the base-2 logarithm (binary logarithm) of `n`.
3. Add 1 to the result of the logarithmic calculation. This is because the logarithm calculation gives the exponent that, when raised to 2, results in `n`. Adding 1 accounts for the leftmost bit (the highest power of 2).
4. Print the original number `n` and the calculated number of bits.

## Resultant Output Explanation

For the given code and the provided test cases, let's break down the output:

1. For the number 7:

• Binary representation: 111
• Calculated bits: log2(7) + 1 = 2.80735492205 + 1 ≈ 3
• The output is "Number: 7\nBits: 3".
2. For the number 16:

• Binary representation: 10000
• Calculated bits: log2(16) + 1 = 4 + 1 = 5
• The output is "Number: 16\nBits: 5".
3. For the number 10:

• Binary representation: 1010
• Calculated bits: log2(10) + 1 ≈ 3.32192809489 + 1 ≈ 4
• The output is "Number: 10\nBits: 4".

## Time Complexity

The time complexity of the algorithm primarily depends on the logarithmic calculation, which is usually a fast operation. The overall complexity is determined by the time complexity of the logarithm function. Assuming that logarithm calculations are constant time (which is generally true for practical purposes), the time complexity of the algorithm is O(1), or constant time.

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