fastest way to modify part of array

Question

I want to perform not operation to a block of continous elements of a bool array and then read back the complete array. I am using the following code to perform the operation.

bool arr[100000]={0};
cin>>x>>y;
for(i=x; i<=y; i++)
 arr[i]=!arr[i];

//Some other operations on the array

for(i=0; i<=100000; i++)
 arr+=arr[i];

This works fine but i am trying to increase the speed of the program. Is there a better way to perform the same operation?

Answer 1

Since I made the comment about using ints (or indeed int64), I may as well write it up and you can evaluate whether it's worth it. It would be something like this. Forgive any errors, as I'm just bunging this into a browser while my kids are watching ridiculously trashy saturday-morning cartoons.

// I'm gonna assume 32-bit ints here.  Makes the other maths clearer.
// Sorry about all the '4' and '32' constants =P
const size_t arrLen = 100000 / 4 + 1;
int arr[arrLen];

//This gets filled with your data...
memset((void*)arr, 0, arrLen*4);

cin >> x >> y;
int leftMask = 0xffffffff >> (x % 32);      // "(x & 0x1f)" faster?
int rightMask = ~(0x7fffffff >> (y % 32));  // "(y & 0x1f)" faster?
x /= 32;                                    // "x >>= 5" faster?
y /= 32;                                    // "y >>= 5" faster?

if( x == y )
{
    // Intersect the masks
    leftMask &= rightMask;
    arr[x] = (arr[x] & ~leftMask) | (~arr[x] & leftMask);
}
else if( x < y )
{
    // Flip the left and right ends
    arr[x] = (arr[x] & ~leftMask) | (~arr[x] & leftMask);
    arr[y] = (arr[y] & ~rightMask) | (~arr[y] & rightMask);

    // Flip everything in between
    for( int i = x+1; i < y; i++ ) {
        arr[i] ^= 0xffffffff;  // Or arr[i] = ~arr[i] -- whichever is faster
    }
}

Alternative for the above loop, if it makes any difference...

// Flip everything in between
for( int *a = arr+x+1, *b = arr+y; a < b; a++ ) {
    *a = ~*a;
}

Exercise is to try with 64-bit integers. Personally, I reckon this approach would be faster than anything else except in the cases where you are only flipping a few bits.

I might have an off-by-one-bit error in the right-hand mask. If anyone spots it please comment. Brain empty. =)

Answer 2

Consider to use bitset. Compare performance - maybe it will be better.

std::bitset<100000> arr;
cin>>x>>y;
for(i=x; i<=y; i++)
 arr.flip(i);

//Some other operations on the array
unsigned int carr = arr.count();

For even more optimized (please measure and don't believe) you can use your own version of bitset<>, THIS IS NOT TESTED CODE:

const size_t arr_bitlen = 100000;
typedef unsigned int arr_type;
const size_t arr_type_size = sizeof(arr_type);
const size_T arr_len = (arr_bitlen + arr_type_size - 1) / arr_type_size;
arr_type arr[arr_len] = { 0 };
cin>>x>>y;
unsigned int x_addr = x / arr_type_size;
unsigned int y_addr = y / arr_type_size;
unsigned int x_bit = x % arr_type_size;
unsigned int y_bit = y % arr_type_size;

if (0 == x_bit)
    for (i=x_addr; i<=y_addr; i++)
       arr[i] = ~arr[i]; // revert all bits (bools)
else {
  // deal with first element in range ( ....xxxx - change only x-s
  arr_type x_mask = ((1 << x_bit) - 1) << (arr_type_len - x_bit);
  arr[x_addr] ^= x_mask; 
  for (i = x_bit + 1; i < arr_type_size; ++i)
      arr[i] = ~arr[i]; // revert all bits (bools)
}
if (y_bit > 0) // try to invert 0..y_bit in arr[y_addr + 1] by yourself

//Some other operations on the array
see implementation of std::bitset<N>::count() - it is very clever - just copy it