Maximum 0’s between two immediate 1’s in binary representation

3.6

Given a number n, the task is to find the maximum 0’s between two immediate 1’s in binary representation of given n. Return -1 if binary representation contains less than two 1’s.

Examples:

Input : n = 47
Output: 1
// binary of n = 47 is 101111

Input : n = 549
Output: 3
// binary of n = 549 is 1000100101

Input : n = 1030
Output: 7
// binary of n = 1030 is 10000000110

Input : n = 8
Output: -1
// There is only one 1 in binary representation
// of 8.

The idea to solve this problem is to use shift operator. We just need to find the position of two immediate 1’s in binary representation of n and maximize the difference of these position.

  • Return -1 if number is 0 or is a power of 2. In these cases there are less than two 1’s in binary representation.
  • Initialize variable prev with position of first right most 1, it basically stores the position of previously seen 1.
  • Now take another variable cur which stores the position of immediate 1 just after prev.
  • Now take difference of cur – prev – 1, it will be the number of 0’s between to immediate 1’s and compare it with previous max value of 0’s and update prev i.e; prev=cur for next iteration.
  • Use auxiliary variable setBit , which scans all bits of n and helps to detect if current bits is 0 or 1.
  • Initially check if N is 0 or power of 2.
// C++ program to find maximum number of 0's
// in binary representation of a number
#include<bits/stdc++.h>
using namespace std;

// Returns maximum 0's between two immediate
// 1's in binary representation of number
int maxZeros(int n)
{
    // If there are no 1's or there is only
    // 1, then return -1
    if (n==0 || (n&(n-1)) == 0)
       return -1;

    // loop to find position of right most 1
    // here sizeof int is 4 that means total 32 bits
    int setBit = 1, prev = 0, i;
    for (i=1; i<=sizeof(int)*8; i++)
    {
         prev++;

         // we have found right most 1
         if ((n & setBit) == setBit)
         {
              setBit = setBit << 1;
              break;
         }

         // left shift setBit by 1 to check next bit
         setBit = setBit << 1;
     }

     // now loop through for remaining bits and find
     // position of immediate 1 after prev
     int max0 = INT_MIN, cur = prev;
     for (int j=i+1; j<=sizeof(int)*8; j++)
     {
         cur++;

         // if cuurent bit is set, then compare
         // difference of cur - prev -1 with
         // previous maximum number of zeros
         if ((n & setBit) == setBit)
         {
              if (max0 < (cur - prev - 1))
                 max0 = cur - prev - 1;

              // update prev
              prev = cur;
         }
         setBit = setBit << 1;
     }
     return max0;
}

// Driver program to run the case
int main()
{
   int n = 549;

   // Initially check that number must not
   // be 0 and power of 2
   cout << maxZeros(n);
   return 0;
}

Output:

3

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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