Baum Sweet Sequence is an infinite binary sequence of 0s and 1s. The nth term of the sequence is 1 if the number n has no odd number of contiguous zeroes in its binary representation, else the nth term is 0.
The first few terms of the sequence are:
b1 = 1 (binary of 1 is 1)
b2 = 0 (binary of 2 is 10)
b3 = 1 (binary of 3 is 11)
b4 = 1 (binary of 4 is 100)
b5 = 0 (binary of 5 is 101)
b6 = 0 (binary of 6 is 110)
Given a natural number n. The task is to find the nth term of the Baum Sweet sequence, i.e, check whether it contains any consecutive block of zeroes of odd length.
Input: n = 8
Output: 0
Explanations:
Binary representation of 8 is 1000. It
contains odd length block of consecutive 0s.
Therefore B8 is 0.
Input: n = 5
Output: 0
Input: n = 7
Output: 1
The idea is to run a loop through the binary representation of n and count the length of all the consecutive zero blocks present. If there is at-least one odd length zero block, then the nth term for the given input n is 0 else it is 1.
CPP
#include <bits/stdc++.h>
using namespace std;
int nthBaumSweetSeq( int n)
{
bitset<32> bs(n);
int len = 32 - __builtin_clz(n);
int baum = 1;
for ( int i = 0; i < len;) {
int j = i + 1;
if (bs[i] == 0) {
int cnt = 1;
for (j = i + 1; j < len; j++) {
if (bs[j] == 0)
cnt++;
else
break ;
}
if (cnt % 2 == 1)
baum = 0;
}
i = j;
}
return baum;
}
int main()
{
int n = 8;
cout << nthBaumSweetSeq(n);
return 0;
}
|
Java
class GFG {
static int nthBaumSweetSeq( int n)
{
char [] bs
= (Integer.toBinaryString(n)).toCharArray();
int baum = 1 ;
for ( int i = 0 ; i < bs.length;) {
int j = i + 1 ;
if (bs[i] == '0' ) {
int cnt = 1 ;
for (j = i + 1 ; j < bs.length; j++) {
if (bs[j] == '0' )
cnt += 1 ;
else
break ;
}
if (cnt % 2 == 1 )
baum = 0 ;
}
i = j;
}
return baum;
}
public static void main(String[] args)
{
int n = 8 ;
System.out.println(nthBaumSweetSeq(n));
}
}
|
Python3
def nthBaumSweetSeq(n):
bs = list ( bin (n)[ 2 ::])
baum = 1
for i in range ( len (bs)):
j = i + 1
if (bs[i] = = '0' ):
cnt = 1
for j in range (i + 1 , len (bs)):
if (bs[j] = = 0 ):
cnt + = 1
else :
break
if (cnt % 2 = = 1 ):
baum = 0
i = j
return baum
n = 8
print (nthBaumSweetSeq(n))
|
C#
using System;
public class GFG {
static int nthBaumSweetSeq( int n)
{
char [] bs
= (Convert.ToString(n, 2)).ToCharArray();
int baum = 1;
for ( int i = 0; i < bs.Length;) {
int j = i + 1;
if (bs[i] == '0' ) {
int cnt = 1;
for (j = i + 1; j < bs.Length; j++) {
if (bs[j] == '0' )
cnt += 1;
else
break ;
}
if (cnt % 2 == 1)
baum = 0;
}
i = j;
}
return baum;
}
public static void Main()
{
int n = 8;
Console.WriteLine(nthBaumSweetSeq(n));
}
}
|
Javascript
function nthBaumSweetSeq(n)
{
let bs = n.toString(2).split( "" );
let baum = 1;
for (let i = 0; i < bs.length;)
{
let j = i + 1;
if (bs[i] == '0' )
{
let cnt = 1;
for (j = i + 1; j < bs.length; j++)
{
if (bs[j] == '0' )
cnt += 1;
else
break ;
}
if (cnt % 2 == 1)
baum = 0;
}
i = j;
}
return baum;
}
let n = 8;
console.log(nthBaumSweetSeq(n));
|
Last Updated :
28 Feb, 2023
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