Given a optimistic integer **N**, the duty is to rely the sum of the variety of set bits within the binary illustration of all of the numbers from **1** to **N**.

**Examples:**

**Input:** N = three**Output:** four

Decimal

Binary

Set Bit Count

1

01

1

2

10

1

three

11

2

1 + 1 + 2 = four

**Input:** N = 16**Output:** 33

**Approach:** Some different approaches to unravel this drawback has been mentioned right here. In this text, one other method with time complexity O(logN) has been mentioned.

Check the sample of Binary illustration of the numbers from 1 to N within the following desk:

Decimal

E

D

C

B

A

zero

zero

zero

zero

zero

zero

1

zero

zero

zero

zero

1

2

zero

zero

zero

1

zero

three

zero

zero

zero

1

1

four

zero

zero

1

zero

zero

5

zero

zero

1

zero

1

6

zero

zero

1

1

zero

7

zero

zero

1

1

1

eight

zero

1

zero

zero

zero

9

zero

1

zero

zero

1

10

zero

1

zero

1

zero

11

zero

1

zero

1

1

12

zero

1

1

zero

zero

13

zero

1

1

zero

1

14

zero

1

1

1

zero

15

zero

1

1

1

1

16

1

zero

zero

zero

zero

Notice that,

Every alternate bits in A are set.

Every 2 alternate bits in B are set.

Every four alternate bits in C are set.

Every eight alternate bits in D are set.

…..

This will carry on repeating for each energy of two.

So, we’ll iterate until the variety of bits within the quantity. And we don’t must iterate each single quantity within the vary from 1 to n.

We will carry out the next operations to get the specified consequence.

, First of all, we’ll add 1 to the quantity with a purpose to compensate zero. As the binary quantity system begins from zero. So now n = n + 1.

We will hold the observe of the variety of set bits encountered until now. And we’ll initialise it with n/2.

We will hold one variable which is an influence of two, with a purpose to hold observe of bit we’re computing.

We will iterate until the ability of two turns into larger than n.

We can get the variety of pairs of 0s and 1s within the present bit for all of the numbers by dividing n by present energy of two.

Now now we have so as to add the bits within the set bits rely. We can do that by dividing the variety of pairs of 0s and 1s by 2 which is able to give us the variety of pairs of 1s solely and after that, we’ll multiply that with the present energy of two to get the rely of ones within the teams.

Now there could also be an opportunity that we get a quantity as variety of pairs, which is someplace in the course of the group i.e. the variety of 1s are lower than the present energy of two in that exact group. So, we’ll discover modulus and add that to the rely of set bits which shall be clear with the assistance of an instance.

**Example:** Consider N = 14

From the desk above, there shall be 28 set bits in complete from 1 to 14.

We shall be contemplating 20 as A, 21 as B, 22 as C and 23 as D

First of all we’ll add 1 to quantity N, So now our N = 14 + 1 = 15.

Calculation for A (20 = 1)

15/2 = 7

Number of set bits in A = 7 ————> (i)

Calculation for B (2^1 = 2)

15/2 = 7 => there are 7 teams of 0s and 1s

Now, to compute variety of teams of set bits solely, now we have to divide that by 2.

So, 7/2 = three. There are three set bit teams.

And these teams will comprise set bits equal to energy of two this time, which is 2. So we’ll multiply variety of set bit teams with energy of two

=> three*2 = 6 —>(2i)

Plus

There could also be some further 1s on this as a result of 4th group shouldn’t be thought of, as this division will give us solely integer worth. So now we have so as to add that as effectively. Note: – This will occur solely when variety of teams of 0s and 1s is odd.

15%2 = 1 —>(2ii)

2i + 2ii => 6 + 1 = 7 ————>(ii)

Calculation for C (2^2 = four)

15/four = three => there are three teams of 0s and 1s

Number of set bit teams = three/2 = 1

Number of set bits in these teams = 1*four = four —> (3i)

As three is odd, now we have so as to add bits within the group which isn’t thought of

So, 15%four = three —> (3ii)

3i + 3ii = four + three = 7 ————>(iii)

Calculation for D (2^three = eight)

15/eight = 1 => there may be 1 group of 0s and 1s. Now on this case there is just one group and that too of solely zero.

Number of set bit teams = half = zero

Number of set bits in these teams = zero * eight = zero —> (4i)

As variety of teams are odd,

So, 15%eight = 7 —> (4ii)

4i + 4ii = zero + 7 = 7 ————>(iv)

At this level, our energy of two variable turns into larger than the quantity, which is 15 in our case. (energy of two = 16 and 16 > 15). So the loop will get terminated right here.

Final output = i + ii + iii + iv = 7 + 7 + 7 + 7 = 28

Number of set bits from 1 to 14 are 28.

Below is the implementation of the above method:

#embrace

utilizing namespace std;

int relySetBits(int n)

int foremost()

int n = 14;

cout << relySetBits(n);

return zero;

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