Check whether bits are in alternate pattern in the given range
Given a non-negative number n and two values l and r. The problem is to check whether or not n has an alternate pattern in its binary representation in the range l to r. Here alternate pattern means that the set and unset bits occur in alternate order. The bits are numbered from right to left, i.e., the least significant bit is considered to be at first position.
Constraint: 1 <= l <= r <= number of bits in the binary representation of n.
Examples:
Input : n = 18, l = 1, r = 3
Output : Yes
(18)10 = (10010)2
The bits in the range 1 to 3 in the binary representation of 18 are in alternate order.
Input : n = 22, l = 2, r = 4
Output : No
(22)10 = (10110)2
The bits in the range 2 to 4 in the binary representation of 22 are not in alternate order.
Algorithm:
bitsAreInAltPatrnInGivenTRange(n, l, r)
// Shift bits of given number n by
// (l-1).
num = n >> (l - 1)
// Find first bit of range
prev = num & 1
// Traverse through remaining
// bits. For every bit, check if
// current bit is same as previous
// or not.
num = num >> 1
for i = 1 to (r - l)
curr = num & 1
if curr == prev then
return false
prev = curr
num = num >> 1
return true;
C++
#include <bits/stdc++.h>
using namespace std;
bool bitsAreInAltPatrnInGivenTRange(unsigned int n,
unsigned int l,
unsigned int r)
{
unsigned int num, prev, curr;
num = n >> (l - 1);
prev = num & 1;
num = num >> 1;
for ( int i = 1; i <= (r - l); i++) {
curr = num & 1;
if (curr == prev)
return false ;
prev = curr;
num = num >> 1;
}
return true ;
}
int main()
{
unsigned int n = 18;
unsigned int l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
class GFG
{
static boolean bitsAreInAltPatrnInGivenTRange( int n,
int l, int r)
{
int num, prev, curr;
num = n >> (l - 1 );
prev = num & 1 ;
num = num >> 1 ;
for ( int i = 1 ; i <= (r - l); i++)
{
curr = num & 1 ;
if (curr == prev)
return false ;
prev = curr;
num = num >> 1 ;
}
return true ;
}
public static void main(String[] args)
{
int n = 18 ;
int l = 1 , r = 3 ;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def bitsAreInAltPatrnInGivenTRange(n, l, r):
num = n >> (l - 1 );
prev = num & 1 ;
num = num >> 1 ;
for i in range ( 1 ,(r - l)):
curr = num & 1 ;
if (curr = = prev):
return False ;
prev = curr;
num = num >> 1 ;
return True ;
if __name__ = = "__main__" :
n = 18 ;
l = 1 ;
r = 3 ;
if (bitsAreInAltPatrnInGivenTRange(n, l, r)):
print ( "Yes" );
else :
print ( "No" );
|
C#
using System;
class GFG
{
static bool bitsAreInAltPatrnInGivenTRange( int n,
int l, int r)
{
int num, prev, curr;
num = n >> (l - 1);
prev = num & 1;
num = num >> 1;
for ( int i = 1; i <= (r - l); i++)
{
curr = num & 1;
if (curr == prev)
return false ;
prev = curr;
num = num >> 1;
}
return true ;
}
static void Main()
{
int n = 18;
int l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
PHP
<?php
function bitsAreInAltPatrnInGivenTRange( $n , $l , $r )
{
$num = $n >> ( $l - 1);
$prev = $num & 1;
$num = $num >> 1;
for ( $i = 1; $i <= ( $r - $l ); $i ++)
{
$curr = $num & 1;
if ( $curr == $prev )
return false;
$prev = $curr ;
$num = $num >> 1;
}
return true;
}
$n = 18;
$l = 1;
$r = 3;
if (bitsAreInAltPatrnInGivenTRange( $n , $l , $r ))
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
function bitsAreInAltPatrnInGivenTRange(n, l, r)
{
var num, prev, curr;
num = n >> (l - 1);
prev = num & 1;
num = num >> 1;
for ( var i = 1; i <= (r - l); i++) {
curr = num & 1;
if (curr == prev)
return false ;
prev = curr;
num = num >> 1;
}
return true ;
}
var n = 18;
var l = 1, r = 3;
if (bitsAreInAltPatrnInGivenTRange(n, l, r))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Output:
Yes
Time complexity: O(r-l)
Auxiliary space: O(1)
Last Updated :
30 Sep, 2022
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