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Code Bit Logic

# Count total set bits in all numbers from 1 to n

The problem we're dealing with is to count the total number of set bits (also known as active bits or 1s) in the binary representation of all numbers from 1 to a given positive integer `n`. This problem involves calculating the sum of set bits in the binary representation of each number within the range.

## Problem Statement and Description

Given a positive integer `n`, we want to count the total number of set bits in the binary representation of all numbers from 1 to `n`. In other words, we need to sum up the number of 1s in the binary representation of each number within the range [1, n].

## Example

Consider the number 7. The binary representations of the numbers from 1 to 7 are as follows:

• 1: 0001 (1 set bit)
• 2: 0010 (1 set bit)
• 3: 0011 (2 set bits)
• 4: 0100 (1 set bit)
• 5: 0101 (2 set bits)
• 6: 0110 (2 set bits)
• 7: 0111 (3 set bits)

The total number of set bits in these binary representations is 12. Therefore, the expected output for this example should be "Output: 12".

## Idea to Solve the Problem

The main idea to solve this problem is to utilize bitwise manipulation to count the number of set bits in the binary representation of each number within the given range. We can use the `countSetBit` function (which you provided) to calculate the number of set bits in a single number. Then, we can iterate through the numbers from 1 to `n` and accumulate the results.

## Pseudocode

Here's the pseudocode for the algorithm:

``````function countSetBit(num):
// Your provided implementation of counting set bits

function countSetBits(n):
result = 0
for i from 1 to n:
result += countSetBit(i)
print "Active Bit from (1 to ", n, ")"
print "Output:", result``````

## Algorithm Explanation

1. `countSetBit(num)` calculates the number of set bits (active bits) in the binary representation of a given number using your provided implementation.
2. `countSetBits(n)` initializes a variable `result` to store the total count of set bits.
3. It then iterates through the numbers from 1 to `n` using a loop.
4. In each iteration, it adds the count of set bits in the binary representation of the current number (`i`) to the `result` variable.
5. After the loop completes, the `result` variable will contain the total number of set bits in the binary representations of all numbers from 1 to `n`.
6. Print the results.

## Time Complexity

The time complexity of the algorithm primarily depends on the iteration through the numbers from 1 to `n` and the calculation of set bits using the `countSetBit` function. Since both of these operations are performed sequentially, the overall time complexity can be approximated as O(n * log n), where `n` is the input number.

## Comment

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