Finding consecutive bit string of 1 or 0

Question

How to find the length of the longest consecutive bit string(either 1 or 0)?

00000000 11110000 00000000 00000000 -> If it is 0 then length will be 20

11111

Answer 1

The following is based on the concept that if you AND a bit sequence with a shifted version of itself, you're effectively removing the trailing 1 from a row of consecutive 1's.

      11101111   (x)
    & 11011110   (x << 1)
    ----------
      11001110   (x & (x << 1)) 
        ^    ^
        |    |
   trailing 1 removed

Repeating this N times will reduce any sequence with N consecutive 1's to 0x00.

So, to count the number of consecutive 1's:

int count_consecutive_ones(int in) {
  int count = 0;
  while (in) {
    in = (in & (in << 1));
    count++;
  }
  return count;
}

To count the number of consecutive 0's, simply invert and the same routine.

int count_consecutive_zeros(int in) {
  return count_consecutive_ones(~in);
} 

---

Proof of concept: http://ideone.com/Z1l0D

int main(void) {
  printf("%d has %d consecutive 1's\n", 0, count_consecutive_ones(0));
  printf("%d has %d consecutive 0's\n", 0, count_consecutive_zeros(0));

  /* 00000000 11110000 00000000 00000000 -> If it is 0 then length will be 20 */
  printf("%x has %d consecutive 0's\n", 0x00F00000, count_consecutive_zeros(0x00F00000));

  /* 11111111 11110000 11110111 11111111 -> If it is 1 then length will be 12 */
  printf("%x has %d consecutive 1's\n", 0xFFF0F7FF, count_consecutive_ones(0xFFF0F7FF));
}

Output:

0 has 0 consecutive 1's
0 has 32 consecutive 0's
f00000 has 20 consecutive 0's
fff0f7ff has 12 consecutive 1's

Answer 2

public static int maxConsecutiveOneInBinaryNumber(int number) {
int count = 0;
int max = 0;
while (number != 0) {
  if ((number & 1) == 1) {
    count++;
  } else {
    max = Math.max(count, max);
    count = 0;
  }
  number = number >> 1;
}
return Math.max(count, max);
}

You can this code here: https://github.com/VishalSKumar/DSFiddle/blob/master/src/main/java/com/vishalskumar/hackerrank/MaxConsecutiveOneInBinary.java

Answer 3

You can form a look up table to do it quickly for you. The bigger the table, the faster the lookup. 2x256 entry tables can do 8 bits at a time with a little bit twiddling. Add a 1s version of the table and start adding entries. That's probably how I'd go about it.

Answer 4

If you're just looking for a byte string of four bytes, you can pack these into an unsigned long and use an algorithm like this:

int CountConsecutiveOnes(unsigned long n)
{
    unsigned long m = n;
    int k = 0;

    while (m)
    {
        ++k;
        n >>= 1;
        m &= n;
    }

    return k;
}

For counting zeros, just taking the bitwise complement first.

If you need to count byte strings longer than four, you can just implement the operations x >>= 1 and x & y either directly on the byte strings or it may be more efficient to use strings of unsigned long so the carry checks on the implementation of x >>= 1 aren't too expensive.

Answer 5

One simple way would be to simply loop over the bits, and keep track of the number of bits in a row which have had the same value, and the maximum that this value has reached.

Here's a simple C function which does just this:

int num_conseq_matching_bits(int n) {
    int i, max, cur, b, prevb;
    prevb = n & 1; /* 0th bit */
    cur = 1;
    max = 1;
    for(i=1; i<32; i++) {
        b = (n >> i) & 1; /* get the i'th bit's value */
        if(b == prevb) {
            cur += 1;
            if(cur > max)
                max = cur;
        }
        else {
            cur = 1; /* count self */
            prevb = b;
        }
    }
    return max;
}

Answer 6

To use the table idea, you need something like

static struct {
    int lead;  /* leading 0 bits */
    int max;   /* maximum 0 bits */
    int trail; /* trailing 0 bits */
} table[256] = { ....data.... };

int mostConsecutiveBits(unsigned char *str, int length, bool count_ones) {
    int max = 0; /* max seen so far */
    int trail = 0; /* trailing 0s from previous bytes */
    while (length-- > 0) {
        int byte = *str++;
        if (count_ones)
            byte ^= 0xff;
        if (table[byte].max > max)
            max = table[byte].max;
        if (trail + table[byte].lead > max)
            max = trail + table[byte].lead;
        if (byte)
            trail = table[byte].trail;
        else
            trail += 8;
    }
    return max;
}

initializing the table is straight-forward, but depends on your bit- and byte-ordering (little endian or big endian).