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# Count unset bits in a range in C++

We are given an integer number let’s say, num and the range with left and right values. The task is to firstly calculate the binary digit of a number and then set the loop from the left digit till the right digit and then in the given range calculate the unset bits.

Unset bits in a binary number is represented by 0. Whenever we calculate the binary number of an integer value then it is formed as the combination of 0’s and 1’s. So, the digit 0 is known as unset bit in the terms of the computer.

**Input** − int number = 50, left = 2, right = 5

**Output** − Count of total unset bits in a range are − 2

**Explanation** −Binary representation of a number 50 is 110010 and we have a range starting from left = 2 which is having bit as 1 and ending to right = 5 which is having bit 1 and in between the range we have two 0’s. So the count of unset bits is 2.

**Input** − int number = 42, left = 1, right 6

**Output** − Count of total unset bits in a range are − 3

**Explanation** − Binary representation of a number 42 is 101010 and we have a range starting from left = 1 which is having bit as 1 and ending to right = 6 which is having bit 0 and in between the range we have three 0’s. So the count is 3.

## Approach used in the below program is as follows

Input the number in a variable of integer type and also the range with left and right integer values.

Declare a variable count to store the total count of set bits of type unsigned int

Start loop FOR from i to 1<<7 and i > 0 and i to i / 2

Inside the loop, check num & 1 == TRUE then print 1 else print 0

Start loop FOR from i to left till right value

Inside the loop, increment the total number of digits between the given range

Start loop while to calculate the total count of bits till number isn’t 0

Inside the loop, set count = count + number & 1 and also set number >>=1

Set a temporary variable let’s say, a with ((1 << right) - 1) ^ ((1 << (left - 1)) - 1);

Also, set count with count & a

In the end, set count as the total number of bits in a range - total number of set bits in a range.

## Example

#include<iostream> using namespace std; //Count total unset bits in a range unsigned int unset_bits(unsigned int number, unsigned int left, unsigned int right){ unsigned int count = 0; unsigned int total_bits = 0; unsigned i; //display the 8-bit number cout<<"8-bit number of "<<number<<" is: "; for (i = 1 << 7; i > 0; i = i / 2){ (number & i)? cout<<"1": cout<<"0"; } //calculate total number of bits in a given range for(i = left; i<=right; i++){ total_bits++; } //calculate the total bits in a number while (number){ count += number & 1; number >>= 1; } //calculate the set bit in a range int a = ((1 << right) - 1) ^ ((1 << (left - 1)) - 1); count = count & a; //subtract set bits from the total bits in a range count = total_bits - count; cout<<"\nCount of total unset bits in a range are: "<<count; } int main(){ unsigned int number = 80; unsigned int left = 1, right = 4; unset_bits(number, left, right); return 0; }

## Output

If we run the above code it will generate the following output −

8-bit number of 80 is: 01010000 Count of total unset bits in a range are: 2

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