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

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.

Below is the implementation of the above idea :

## C++

`// 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; ` `} ` |

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## Java

`// Java program to find maximum number of 0's ` `// in binary representation of a number ` `class` `GFG { ` ` ` ` ` `// Returns maximum 0's between two immediate ` ` ` `// 1's in binary representation of number ` ` ` `static` `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` `; ` ` ` `} ` ` ` `//int size in java is 4 byte ` ` ` `byte` `b = ` `4` `; ` ` ` `// 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 <= b* ` `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 = Integer.MIN_VALUE, cur = prev; ` ` ` `for` `(` `int` `j = i + ` `1` `; j <= b * ` `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 ` ` ` `static` `public` `void` `main(String[] args) { ` ` ` `int` `n = ` `549` `; ` ` ` ` ` `// Initially check that number must not ` ` ` `// be 0 and power of 2 ` ` ` `System.out.println(maxZeros(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

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## C#

`// C# program to find maximum number of 0's ` `// in binary representation of a number ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// Returns maximum 0's between two immediate ` ` ` `// 1's in binary representation of number ` ` ` `static` `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` `.MinValue, 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 ` ` ` `static` `public` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 549; ` ` ` ` ` `// Initially check that number must not ` ` ` `// be 0 and power of 2 ` ` ` `Console.WriteLine(maxZeros(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by vt_m. ` |

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**Output:**

3

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