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.
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.
Below is the implementation of the above idea :
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