# Maximum 0’s between two immediate 1’s in binary representation in C++

C++Server Side ProgrammingProgramming

## Problem statement

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

## Example

If input number is 35 then its binary representation is −

00100011

In above binary representation there are 3 0’s between two immediate 1’s. Hence answer is 3.

## Algorithm

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

• If number is 0 or power of 2 then return -1
• IInitialize variable prev with position of first right most 1. It stores the position of previously seen 1.
• Take another variable cur which stores the position of immediate 1 just after prev.
• ITake 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.
• IUse variable setBit, which scans all bits of n and helps to detect if current bits is 0 or 1. Use auxiliary variable setBit, which scans all bits of n and helps to detect if current bits is 0 or 1.

## Example

Let us now see an example −

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMaxZeros(int n) {
if (n == 0 || (n & (n - 1) == 0)) {
return -1;
}
int setBit = 1;
int prev = 0;
int i;
for (i = 1; i < sizeof(int) * 8; ++i) {
++prev;
if ((n & setBit) == setBit) {
setBit = setBit << 1;
break;
}
setBit = setBit << 1;
}
int maxZeros = INT_MIN;
int cur = prev;
for (int j = i + 1; j <= sizeof(int) * 8; ++j) {
++cur;
if ((n & setBit) == setBit) {
if (maxZeros < (cur - prev - 1)) {
maxZeros = cur - prev - 1; prev = cur;
}
}
setBit = setBit << 1;
}
return maxZeros;
}
int main() {
int n = 35;
cout << "Maximum zeros = " << getMaxZeros(n) << endl;
return 0;
}

## Output

Maximum zeros = 3
Published on 31-Dec-2019 11:52:24