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

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