# 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.
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## Python3

 `# Python3 program to find maximum number of  ` `# 0's in binary representation of a number ` ` `  `# Returns maximum 0's between two immediate ` `# 1's in binary representation of number ` `def` `maxZeros(n): ` `    ``# If there are no 1's or there is  ` `    ``# only 1, then return -1 ` `    ``if` `(n ``=``=` `0` `or` `(n & (n ``-` `1``)) ``=``=` `0``): ` `        ``return` `-``1` ` `  `    ``# loop to find position of right most 1 ` `    ``# here sizeof is 4 that means total 32 bits ` `    ``setBit ``=` `1` `    ``prev ``=` `0` `    ``i ``=` `1` `    ``while``(i < ``33``): ` `        ``prev ``+``=` `1` ` `  `        ``# 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 ` `    ``max0 ``=` `-``10``*``*``9` `    ``cur ``=` `prev ` `    ``for` `j ``in` `range``(i ``+` `1``, ``33``): ` `        ``cur ``+``=` `1` ` `  `        ``# 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 Code ` `n ``=` `549` ` `  `# Initially check that number must not ` `# be 0 and power of 2 ` `print``(maxZeros(n)) ` ` `  `# This code is contributed by Mohit Kumar `

## 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. `

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.