Check if bits of a number has count of consecutive set bits in increasing order
Last Updated :
11 Sep, 2023
Given a integer n > 0, the task is to find whether in the bit pattern of integer count of continuous 1’s are in increasing from left to right.
Examples :
Input:19
Output:Yes
Explanation: Bit-pattern of 19 = 10011,
Counts of continuous 1's from left to right
are 1, 2 which are in increasing order.
Input : 183
Output : yes
Explanation: Bit-pattern of 183 = 10110111,
Counts of continuous 1's from left to right
are 1, 2, 3 which are in increasing order.
A simple solution is to store binary representation of given number into a string, then traverse from left to right and count the number of continuous 1’s. For every encounter of 0 check the value of previous count of continuous 1’s to that of current value, if the value of previous count is greater than the value of current count then return False, Else when string ends return True.
C++
#include <bits/stdc++.h>
using namespace std;
bool findContinuous1( int n)
{
const int bits = 8 * sizeof ( int );
string bp = bitset<bits>(n).to_string();
int prev_count = 0, curr_count = 0;
int i = 0;
while (i < bits) {
if (bp[i] == '1' ) {
curr_count++;
i++;
}
else if (bp[i - 1] == '0' ) {
i++;
curr_count = 0;
continue ;
}
else {
if (curr_count < prev_count)
return 0;
i++;
prev_count = curr_count;
curr_count = 0;
}
}
if (prev_count > curr_count && (curr_count != 0))
return 0;
return 1;
}
int main()
{
int n = 179;
if (findContinuous1(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
import java.io.*;
public class GFG {
static boolean findContinuous1( int n)
{
char [] bp
= (Integer.toBinaryString(n)).toCharArray();
int bits = bp.length;
int prev_count = 0 ;
int curr_count = 0 ;
int i = 0 ;
while (i < bits) {
if (bp[i] == '1' )
{
curr_count++;
i++;
}
else if (bp[i - 1 ] == '0' ) {
i++;
curr_count = 0 ;
continue ;
}
else {
if (curr_count < prev_count)
return false ;
i++;
prev_count = curr_count;
curr_count = 0 ;
}
}
if ((prev_count > curr_count) && (curr_count != 0 ))
return false ;
return true ;
}
public static void main(String args[])
{
int n = 179 ;
if (findContinuous1(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def findContinuous1(n):
bp = list ( bin (n))
bits = len (bp)
prev_count = 0
curr_count = 0
i = 0
while (i < bits):
if (bp[i] = = '1' ):
curr_count + = 1
i + = 1
elif (bp[i - 1 ] = = '0' ):
i + = 1
curr_count = 0
continue
else :
if (curr_count < prev_count):
return 0
i + = 1
prev_count = curr_count
curr_count = 0
if (prev_count > curr_count and (curr_count ! = 0 )):
return 0
return 1
n = 179
if (findContinuous1(n)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
using System.Collections.Specialized;
class GFG
{
static bool findContinuous1( int n)
{
string bp = Convert.ToString(n, 2);
int bits = bp.Length;
int prev_count = 0, curr_count = 0;
int i = 0;
while (i < bits)
{
if (bp[i] == '1' )
{
curr_count++;
i++;
}
else if (bp[i - 1] == '0' ) {
i++;
curr_count = 0;
continue ;
}
else {
if (curr_count < prev_count)
return false ;
i++;
prev_count = curr_count;
curr_count = 0;
}
}
if (prev_count > curr_count && (curr_count != 0))
return false ;
return true ;
}
static void Main()
{
int n = 179;
if (findContinuous1(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
Javascript
<script>
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
function findContinuous1(n){
let bp = dec2bin(n)
let bits = bp.length
let prev_count = 0
let curr_count = 0
let i = 0
while (i < bits){
if (bp[i] == '1' ){
curr_count += 1;
i += 1;
}
else if (bp[i - 1] == '0' ){
i += 1
curr_count = 0
continue
}
else {
if (curr_count < prev_count)
return 0
i += 1
prev_count = curr_count
curr_count = 0
}
}
if (prev_count > curr_count && (curr_count != 0))
return 0
return 1
}
n = 179
if (findContinuous1(n))
document.write( "Yes" )
else
document.write( "No" )
</script>
|
Time Complexity: O(logn)
Auxiliary Space: O(logn)
An efficient solution is to use decimal to binary conversion loop that divides number by 2 and take remainder as bit. This loop finds bits from right to left. So we check if right to left is in decreasing order or not.
Below is the implementation.
C++
#include<bits/stdc++.h>
using namespace std;
bool areSetBitsIncreasing( int n)
{
int prev_count = INT_MAX;
while (n > 0)
{
while (n > 0 && n % 2 == 0)
n = n/2;
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}
if (curr_count >= prev_count)
return false ;
prev_count = curr_count;
}
return true ;
}
int main()
{
int n = 10;
if (areSetBitsIncreasing(n))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
import java .io.*;
class GFG {
static boolean areSetBitsIncreasing( int n)
{
int prev_count = Integer.MAX_VALUE;
while (n > 0 )
{
while (n > 0 && n % 2 == 0 )
n = n/ 2 ;
int curr_count = 1 ;
while (n > 0 && n % 2 == 1 )
{
n = n/ 2 ;
curr_count++;
}
if (curr_count >= prev_count)
return false ;
prev_count = curr_count;
}
return true ;
}
static public void main (String[] args)
{
int n = 10 ;
if (areSetBitsIncreasing(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
import sys
def areSetBitsIncreasing(n):
prev_count = sys.maxsize
while (n > 0 ):
while (n > 0 and n % 2 = = 0 ):
n = int (n / 2 )
curr_count = 1
while (n > 0 and n % 2 = = 1 ):
n = n / 2
curr_count + = 1
if (curr_count > = prev_count):
return False
prev_count = curr_count
return True
n = 10
if (areSetBitsIncreasing(n)):
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool areSetBitsIncreasing( int n)
{
int prev_count = int .MaxValue;
while (n > 0)
{
while (n > 0 && n % 2 == 0)
n = n/2;
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}
if (curr_count >= prev_count)
return false ;
prev_count = curr_count;
}
return true ;
}
static public void Main ()
{
int n = 10;
if (areSetBitsIncreasing(n))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
PHP
<?php
function areSetBitsIncreasing( $n )
{
$prev_count = PHP_INT_MAX;
while ( $n > 0)
{
while ( $n > 0 && $n % 2 == 0)
$n = $n / 2;
$curr_count = 1;
while ( $n > 0 and $n % 2 == 1)
{
$n = $n / 2;
$curr_count ++;
}
if ( $curr_count >= $prev_count )
return false;
$prev_count = $curr_count ;
}
return true;
}
$n = 10;
if (areSetBitsIncreasing( $n ))
echo "Yes" ;
else
echo "No" ;
?>
|
Javascript
<script>
function areSetBitsIncreasing(n)
{
var prev_count = Number.MAX_VALUE;
while (n > 0)
{
while (n > 0 && n % 2 == 0)
n = parseInt(n / 2);
var curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n / 2;
curr_count++;
}
if (curr_count >= prev_count)
return false ;
prev_count = curr_count;
}
return true ;
}
var n = 10;
if (areSetBitsIncreasing(n))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(log2n)
Auxiliary Space: O(1)
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