Discussion on Consecutive 1's in Binary Numbers Challenge

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6 years ago+ 11 comments

The problem can be solved most efficiently using bit-manipulation. Once again the editorial disappointed me. "Discussions" has better solutions than the editorial. Anyways, the problem can be solved in O(b) complexity, where b=consecutive 1's :

n = int(input())
ans = 0
while n>0 :
    n &= (n<<1)
    ans += 1
print(ans)

Logic :

"Bitwise-and" accompanied with "left-shift" reduces consecutive set-bits(1's) by 1. Thus, counting these operations gives the required answer.

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6 years ago+ 3 comments

public class Solution {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int count=0;
        int result=0;
        while(n!=0)
            {
            if(n%2==1)
                {
               count++;
               result=fun(count,result);
            }
            else
                {
               result=fun(count,result);
                count=0;
            }
            n=n/2;
        }
        System.out.println(result);       
}
    public static int fun(int a,int b)
        {
        if(b<a)
          b=a;
        return b;
    }
}

2 years ago+ 0 comments

this problem solution can be optimize further. here is the trick to solve this problem in O(K).

https://youtu.be/bc7cxeDy308

2 years ago+ 2 comments

Those who are looking for javascript solution

const n = parseInt(readLine(), 10);
    
    let count = 0;
    let result = 0;
    let arr = n.toString(2);
    for(let i = 0; i< arr.length; i++) {
        if(arr[i] == 0) {
            count = 0;
        } else {
            count++;
            result = Math.max(result, count)
        }

    }
    console.log(result)

3 years ago+ 1 comment

Stack stack=new Stack(); int count=1,i=-1,max=0; while(n!=0) { stack.push(n%2); n=n/2; } while(!stack.isEmpty()) { if(stack.peek()==i) { count++; }

        if(max<count)
        {
            max=count;
        }
        i=stack.peek();

        if(i==0)
        {
            count=1;
        }
        stack.pop();
    }

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