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

It may help you.... First convert your binary number to String say bits. It will give you max number of consecutive 1's (in java)

String[] split = bits.split("0");
Arrays.sort(split);
int maxLength = split[split.length - 1].length();

Answer 2

Posting from iPhone withbig fingers.

If ones, then invert.

Loop over the input using a leadz function. For each iteration, shift the input to the left. Continue until you reach the end of the input. Note that you need to compare the original input length with the cumulative leadz counts.

Also, as an optimization, you can early abort when the remaining input length is less than the largest leadz you have seen.

There are many fast leadz algorithms online.

Answer 3

Since you didn't wrote what is bit string (regular int, byte array or char string I've assumed that it's char array

int maxConsBits(char *pStr,char cChar)
{
    char curChar;
    int curMax = 0;
    int max = 0;
    while (pStr)
    {
       if (*pStr == cChar)
       {
          curMax++;
          if (curMax > max)
          {
             max = curMax;
          }
       }
       else
       {        
           curMax = 0;
       }
       pStr++;
   }
   return max;
}

Answer 4

I don't agree with the tables idea, because I was trying it and realized that even though "BA" in ASCII would contain 5 consecutive 0's for 'B' and 5 consecutive 0's for 'A', they will not add together for 10 consecutive 0's. As a matter of fact, there would be 5 consecutive 0's maximum. (This was in reference to a simple "counting bits in a table idea." Chris Dodd has since expounded on how a table could be used accurately.)

I would use an algorithm like this:

#include <iostream>
#include <algorithm>

using namespace std;

// Assumes Little Endian architecture
int mostConsecutiveBits(char str[], int length) {
    int currentConsecutiveBits=0; 
    int maxConsecutiveBits=0; 
    char currentBit;
    char lastBit=0; 
    char currentChar=str[0];
    int charCtr,bitCtr; 

    for (charCtr=length-1; charCtr>=0; charCtr--) {

        currentChar=str[charCtr];

        for (bitCtr=0; bitCtr<8; bitCtr++) {
            currentBit=currentChar & 1;

            if (currentBit!=lastBit) {
                maxConsecutiveBits=max(maxConsecutiveBits,currentConsecutiveBits);
                currentConsecutiveBits=1; 
                lastBit=currentBit;
            }
            else {
                currentConsecutiveBits++;
            }

            currentChar=currentChar>>1; 

        }
        maxConsecutiveBits=max(maxConsecutiveBits,currentConsecutiveBits);
    }

    return maxConsecutiveBits; 
}   


int main (int argc, char * const argv[]) {
    cout << mostConsecutiveBits("AB",2);
    return 0;
}

In this algorithm, I assume the bitstream is represented as 8-bit characters. For each character, I look at the very last bit with a bitwise AND. If it's the same as the last bit, then I up the consecutive bit count, otherwise, I reset the count because the bits are no longer consecutive. I then use a bitwise shift operation to move the next bit in the character over for observation. Hope this helps!

My answer is effectively a duplicate of David Underhill's answer. :)