<Codesignal> Range Bit Count

Easy

You are given two numbers a and b where 0 ≤ a ≤ b. Imagine you construct an array of all the integers from a to b inclusive. You need to count the number of 1s in the binary representations of all the numbers in the array.

Example

For a = 2 and b = 7, the output should be
rangeBitCount(a, b) = 11.

Given a = 2 and b = 7 the array is: [2, 3, 4, 5, 6, 7]. Converting the numbers to binary, we get [10, 11, 100, 101, 110, 111], which contains 1 + 2 + 1 + 2 + 2 + 3 = 11 1s.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] integer a

  Guaranteed constraints:
  0 ≤ a ≤ b.

• [input] integer b

  Guaranteed constraints:
  a ≤ b ≤ 10.

• [output] integer

[C] Syntax Tips

// Prints help message to the console
// Returns a string
char * helloWorld(char * name) {
    char * answer = malloc(strlen(name) + 8);
    printf("This prints to the console when you Run Tests");
    strcpy(answer, "Hello, ");
    strcat(answer, name);
    return answer;
}

---

Solution

int rangeBitCount(int a,int b)
{
	int count=0;

	for(int i=a;i<=b;i++)
		for(int j=0;j<5;j++)
			count+=(i&(1<<j))!=0;

	return count;
}