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)