I don't think that there is a shortcut, if the array contains
arbitrary data.
You'll want to create an array of bit counts for either 8 or 16 bits.
I would go with 16.
int bitCount[65536];
// Do this once at initialization time
void init()
{
bitCount[0] = 0;
for(int i = 1; i < 65536; ++i) bitCount[i] = bitCount[i/2] + (i&1);
}
// Return the number of bits set in a[n]
int countBits(unsigned int *a, int n)
{
int result = 0;
short *p = (short *)a;
short *q = p + 2*n;
while(p < q)
result += bitCount[*(p++)];
return result;
}
On May 11, 3:35 pm, rahul sharma <rahul23111...@gmail.com> wrote:
> I was searching for google questions and got this question.Use look up to
> do it in bext way
>
> What is best time complexity for this..
> plz post algo too
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