Simplest way to calculate amount of even numbers in given range

Question

What is the simplest way to calculate the amount of even numbers in a range of unsigned integers?

An example: if range is [0...4] then the answer is 3 (0,2,4)

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Answer 1

Hint 1: The modulo operator will return the remainder of the current number
Hint 2: You don't need a for loop
Hint 3: A range is continuous
Hint 4: The number of even numbers in a continuous range are half even (sometimes half + 1, sometimes half - 1)
Hint 5: Building on Hint1: Consider also what (being + end + 1) % 2 gives
Hint 6: Most or all of the answers in this thread are wrong
Hint 7: Make sure you try your solution with negative number ranges
Hint 8: Make sure you try your solution with ranges spanning both negative and positive numbers

Answer 2

int even = (0 == begin % 2) ? (end - begin) / 2 + 1 : (end - begin + 1) / 2;

Which can be converted into:

int even = (end - begin + (begin % 2)) / 2 + (1 - (begin % 2));

EDIT: This can further simplified into:

int even = (end - begin + 2 - (begin % 2)) / 2;

EDIT2: Because of the in my opinion somewhat incorrect definition of integer division in C (integer division truncates downwards for positive numbers and upwards for negative numbers) this formula won't work when begin is a negative odd number.

EDIT3: User 'iPhone beginner' correctly observes that if begin % 2 is replaced with begin & 1 this will work correctly for all ranges.

Answer 3

Answer is to use binary AND.

so a number is represented in memory in 0 and 1. lets say 4 and 5.

4 = 0000 0100

5 = 0000 0101

and every even number has a zero in the end and every odd number has 1 in the end;

in c '1' means true and '0' means false.

so: lets code;

function isEven(int num){
     return ((num & 0x01) == 0) ? 1 : 0;
}

Here 0x01 means 0000 0001. so we are anding 0x01 with the given number.

imagine no is 5

5    |0000 0101 

0x01 |0000 0001

---------------

      0000 0001

so answer will be '1'.

imagine no is 4

4    |0000 0100 

0x01 |0000 0001

---------------

      0000 0000

so answer will be '0'.

now,

return ((num & 0x01) == 0) ? 1 : 0;

it is expanded in :

if((num & 0x01) == 0){// means  the number is even
      return 1;
}else{//means no is odd
      return 0;
}

So this is all.

End is binary operatiors are very important in compititive programming world.

happy coding.

first answer here.

EDIT 1:

Total no of evens are

totalEvens = ((end - start) / 2 + ((((end - start) & 0x01 ) == 0) ? 0 : 1 ));

here (end - start)/2 gives the half of total numbers.

this works if one is even and one is odd.

but,

((((end - start) & 0x01 ) == 0) ? 0 : 1 )

can just replaced by (!isEven(end-start))

So, if the total number is even then dont add 1 else add 1.

this completely works.

Answer 4

The range is always [2a+b, 2c+d] with b,d = {0,1}. Make a table:

b d | #even
0 0 | c-a+1
0 1 | c-a+1
1 0 | c-a
1 1 | c-a+1

Now a = min/2, b = min % 2, c = max/2 and d = max % 2.

So int nEven = max/2 - min/2 + 1 - (min%2).

Answer 5

In terms of start and length:

(length >> 1) + (1 & ~start & length)

half the length plus 1 if start is even and length is odd.

In terms of start and end:

((end - start + 1) >> 1) + (1 & ~start & ~end)

half the length plus 1 if start is even and end is even.

Answer 6

I'm a bit surprised that iteration was tried to solve this. The minimum number of even numbers possible in a range is equal to half of the length of the array of numbers, or, rangeEnd - rangeStart.
Add 1 if the first or last number is even.

So the method is: (using javascript)

function evenInRange(rangeStart, rangeEnd)
{
  return
    Math.floor(rangeEnd - rangeStart) + 
    ((rangeStart % 2 == 0) || (rangeEnd % 2 == 0) ? 1 : 0)
}


Tests:
0 1 2 3 4 5 6 7 8
8 - 0 = 8
8 / 2 = 4
4 + 1 = 5
Even numbers in range:
0 2 4 6 8

11 12 13 14 15 16 17 18 19 20
20 - 11 = 9
9 / 2 = 4
4 + 1 = 5
Even numbers in range
12 14 16 18 20

1 2 3
3 - 1 = 2
2 / 2 = 1
1 + 0 = 1
Even numbers in range
2

2 3 4 5
5 - 2 = 3
3 / 2 = 1
1 + 1 = 2
Even numbers in range
2 4

2 3 4 5 6
6 - 2 = 4
4 / 2 = 2
2 + 1 = 3
Even numbers in range
2 4 6