# 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 current 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 current 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 current 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 current 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.` |

## Javascript

`<script>` `// JavaScript 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` `function` `maxZeros(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` ` ` `let b = 4;` ` ` ` ` `// Loop to find position of right most 1` ` ` `// here sizeof int is 4 that means total 32 bits` ` ` `let 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` ` ` `let max0 = Number.MIN_VALUE, cur = prev;` ` ` `for` `(let j = i + 1; j <= b * 8; j++)` ` ` `{` ` ` `cur++;` ` ` `// If current 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` `let n = 549;` `// Initially check that number must not` `// be 0 and power of 2` `document.write(maxZeros(n));` `// This code is contributed by code_hunt` `</script>` |

**Output:**

3

**Time Complexity**: O(1), because it is taking constant time irrespective of the input n.

**Auxiliary Space**: O(1)

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