We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Tutorials
  3. 30 Days of Code
  4. Day 29: Bitwise AND
  5. Discussions

Day 29: Bitwise AND

Sort 399 Discussions, By:

  • aquajim03
     + 0 comments

    for one liner gang !!!!!!!!!!!!!!!!!!!!!!!!!!

    def bitwiseAnd(N, K):
    return K - 1 if (K - 1 | K <= N) else K - 2

  • dashrek1
     + 0 comments

    In C++ was easy, but in python better dont use itertools combinations only double while.

  • raviiiraj
     + 0 comments

    Java

    Here is a very simple solution with the help of JAVA

    class Result {

    /*
     * Complete the 'bitwiseAnd' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts following parameters:
     *  1. INTEGER N
     *  2. INTEGER K
     */

    public static int bitwiseAnd(int N, int K) {
        int cont=0;
        for(int i=1; i<=N; i++){            //outer loop      
							for(int j=i+1; j<=N; j++){     //inner loop
                if((i&j) < K){  //condition to check the BITAND less then the //the given number
                    if((i&j)>cont){     //to compare max and store it
                        cont = (i&j);
                    }
                }
            }
        }
        return cont;
    }

}

  • raviiiraj
     + 0 comments

    Java

    Here is a very simple solution with the help of JAVA

    class Result {

    /*
     * Complete the 'bitwiseAnd' function below.
     *
     * The function is expected to return an INTEGER.
     * The function accepts following parameters:
     *  1. INTEGER N
     *  2. INTEGER K
     */

    public static int bitwiseAnd(int N, int K) {
        int cont=0;
        for(int i=1; i<=N; i++){         //outer loop
		    		for(int j=i+1; j<=N; j++){     //inner loop
                if((i&j) < K){  //condition to check the BITAND less then the 
								                     //the given number
                    if((i&j)>cont){  //to compare max and store it
                        cont = (i&j);
                    }
                }
            }
        }
        return cont;
    }

}

  • for_andrey_lucky
     + 0 comments

    Python 3

    TL;DR

    def bitwiseAnd(N, K):
    while K > 0:
        K -= 1
        tested = K+(1<<list(reversed(f'{K:b}')).index('0')) if "0" in f"{K:b}" else K+(1<<len(f'{K:b}'))
        if tested <= N :
            return K

    Pros: get solution in max 2 iterations (if I'm not wrong)

    In details

    As example: given K=180. (K-1) = 179 = b10110011

    Find int X, that X & 179 = 179:

    • K-1= 179: b10110011
    • X1 = 183: b10110111
    • X2 = 191: b10111111
    • X3 = 255: b11111111
    • and
    • X4 = 511: b111111111 (note additional leading 1)

    0 is inside binary (K-1). Replace 0 by 1. In most cases binary representation of K-1 has few 0 inside, like in our example: (K-1) = 179 is b10110011. In such cases take 1st candidate 183 b10110111. Test if 183 is less or equal N. If True, then 179 is the answer. If False, no need to test other candidates (191,255,511), because they are defenetly greater that N.

    No 0 is inside binary (K-1). Add leading 1 to binary representation. This is edge case, when binary representation of K-1 is all 1 only. Exampe: K=256, K-1=255: b11111111 (8x1). In this case you need to test if 511: b111111111 (9x1) is less or equal N. If True then 255 is the answer.

    If above test yield False, reduce N by 1 and repeat the test.

    Happy pythoning!

    🦁

Load more conversations