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++
// C++ program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
#include <bits/stdc++.h>
using namespace std;
// Returns true if n has increasing count of
// continuous-1 else false
bool findContinuous1(int n)
{
const int bits = 8 * sizeof(int);
// store the bit-pattern of n into
// bit bitset- bp
string bp = bitset<bits>(n).to_string();
// set prev_count = 0 and curr_count = 0.
int prev_count = 0, curr_count = 0;
int i = 0;
while (i < bits) {
if (bp[i] == '1') {
// increment current count of continuous-1
curr_count++;
i++;
}
// traverse all continuous-0
else if (bp[i - 1] == '0') {
i++;
curr_count = 0;
continue;
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else {
if (curr_count < prev_count)
return 0;
i++;
prev_count = curr_count;
curr_count = 0;
}
}
// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return 0;
return 1;
}
// Driver code
int main()
{
int n = 179;
if (findContinuous1(n))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
import java.io.*;
public class GFG {
// Returns true if n has increasing count of
// continuous-1 else false
static boolean findContinuous1(int n)
{
// converting decimal integer to binary string then
// store the bit-pattern of n into
// bit bitset- bp by converting the binary string to
// char array
char[] bp
= (Integer.toBinaryString(n)).toCharArray();
int bits = bp.length;
// set prev_count = 0 and curr_count = 0.
int prev_count = 0;
int curr_count = 0;
int i = 0;
while (i < bits) {
if (bp[i] == '1')
{
// increment current count of continuous-1
curr_count++;
i++;
}
// traverse all continuous-0
else if (bp[i - 1] == '0') {
i++;
curr_count = 0;
continue;
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else {
if (curr_count < prev_count)
return false;
i++;
prev_count = curr_count;
curr_count = 0;
}
}
// check for last sequence of continuous-1
if ((prev_count > curr_count) && (curr_count != 0))
return false;
return true;
}
// Driver code
public static void main(String args[])
{
int n = 179;
if (findContinuous1(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by phasing17
# Python3 program to find if bit-pattern
# of a number has increasing value of
# continuous-1 or not.
# Returns true if n has increasing count of
# continuous-1 else false
def findContinuous1(n):
# store the bit-pattern of n into
# bit bitset- bp
bp = list(bin(n))
bits = len(bp)
# set prev_count = 0 and curr_count = 0.
prev_count = 0
curr_count = 0
i = 0
while (i < bits):
if (bp[i] == '1'):
# increment current count of continuous-1
curr_count += 1
i += 1
# traverse all continuous-0
elif (bp[i - 1] == '0'):
i += 1
curr_count = 0
continue
# check prev_count and curr_count
# on encounter of first zero after
# continuous-1s
else:
if (curr_count < prev_count):
return 0
i += 1
prev_count = curr_count
curr_count = 0
# check for last sequence of continuous-1
if (prev_count > curr_count and (curr_count != 0)):
return 0
return 1
# Driver code
n = 179
if (findContinuous1(n)):
print("Yes")
else:
print("No")
# This code is contributed by SHUBHAMSINGH10
// C# program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
using System;
using System.Collections.Specialized;
class GFG
{
// Returns true if n has increasing count of
// continuous-1 else false
static bool findContinuous1(int n)
{
// store the bit-pattern of n into
// bit bitset- bp
string bp = Convert.ToString(n, 2);
int bits = bp.Length;
// set prev_count = 0 and curr_count = 0.
int prev_count = 0, curr_count = 0;
int i = 0;
while (i < bits)
{
if (bp[i] == '1')
{
// increment current count of continuous-1
curr_count++;
i++;
}
// traverse all continuous-0
else if (bp[i - 1] == '0') {
i++;
curr_count = 0;
continue;
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else {
if (curr_count < prev_count)
return false;
i++;
prev_count = curr_count;
curr_count = 0;
}
}
// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return false;
return true;
}
// Driver Code
static void Main()
{
int n = 179;
if (findContinuous1(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This Code was contributed by phasing17
<script>
// JavaScript program to find if bit-pattern
// of a number has increasing value of
// continuous-1 or not.
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
// Returns true if n has increasing count of
// continuous-1 else false
function findContinuous1(n){
// store the bit-pattern of n into
// bit bitset- bp
let bp = dec2bin(n)
let bits = bp.length
// set prev_count = 0 and curr_count = 0.
let prev_count = 0
let curr_count = 0
let i = 0
while (i < bits){
if (bp[i] == '1'){
// increment current count of continuous-1
curr_count += 1;
i += 1;
}
// traverse all continuous-0
else if (bp[i - 1] == '0'){
i += 1
curr_count = 0
continue
}
// check prev_count and curr_count
// on encounter of first zero after
// continuous-1s
else{
if (curr_count < prev_count)
return 0
i += 1
prev_count = curr_count
curr_count = 0
}
}
// check for last sequence of continuous-1
if (prev_count > curr_count && (curr_count != 0))
return 0
return 1
}
// Driver code
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++
// C++ program to check if counts of consecutive
// 1s are increasing order.
#include<bits/stdc++.h>
using namespace std;
// Returns true if n has counts of consecutive
// 1's are increasing order.
bool areSetBitsIncreasing(int n)
{
// Initialize previous count
int prev_count = INT_MAX;
// We traverse bits from right to left
// and check if counts are decreasing
// order.
while (n > 0)
{
// Ignore 0s until we reach a set bit.
while (n > 0 && n % 2 == 0)
n = n/2;
// Count current set bits
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}
// Compare current with previous and
// update previous.
if (curr_count >= prev_count)
return false;
prev_count = curr_count;
}
return true;
}
// Driver code
int main()
{
int n = 10;
if (areSetBitsIncreasing(n))
cout << "Yes";
else
cout << "No";
return 0;
}
// Java program to check if counts of
// consecutive 1s are increasing order.
import java .io.*;
class GFG {
// Returns true if n has counts of
// consecutive 1's are increasing
// order.
static boolean areSetBitsIncreasing(int n)
{
// Initialize previous count
int prev_count = Integer.MAX_VALUE;
// We traverse bits from right to
// left and check if counts are
// decreasing order.
while (n > 0)
{
// Ignore 0s until we reach
// a set bit.
while (n > 0 && n % 2 == 0)
n = n/2;
// Count current set bits
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}
// Compare current with previous
// and update previous.
if (curr_count >= prev_count)
return false;
prev_count = curr_count;
}
return true;
}
// Driver code
static public void main (String[] args)
{
int n = 10;
if (areSetBitsIncreasing(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by anuj_67.
# Python3 program to check if counts of
# consecutive 1s are increasing order.
import sys
# Returns true if n has counts of
# consecutive 1's are increasing order.
def areSetBitsIncreasing(n):
# Initialize previous count
prev_count = sys.maxsize
# We traverse bits from right to
# left and check if counts are
# decreasing order.
while (n > 0):
# Ignore 0s until we reach a
# set bit.
while (n > 0 and n % 2 == 0):
n = int(n/2)
# Count current set bits
curr_count = 1
while (n > 0 and n % 2 == 1):
n = n/2
curr_count += 1
# Compare current with previous
# and update previous.
if (curr_count >= prev_count):
return False
prev_count = curr_count
return True
# Driver code
n = 10
if (areSetBitsIncreasing(n)):
print("Yes")
else:
print("No")
# This code is contributed by Smitha
// C# program to check if counts of
// consecutive 1s are increasing order.
using System;
class GFG {
// Returns true if n has counts of
// consecutive 1's are increasing
// order.
static bool areSetBitsIncreasing(int n)
{
// Initialize previous count
int prev_count = int.MaxValue;
// We traverse bits from right to
// left and check if counts are
// decreasing order.
while (n > 0)
{
// Ignore 0s until we reach
// a set bit.
while (n > 0 && n % 2 == 0)
n = n/2;
// Count current set bits
int curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n/2;
curr_count++;
}
// Compare current with previous
// and update previous.
if (curr_count >= prev_count)
return false;
prev_count = curr_count;
}
return true;
}
// Driver code
static public void Main ()
{
int n = 10;
if (areSetBitsIncreasing(n))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by anuj_67.
<?php
// PHP program to check if
// counts of consecutive
// 1s are increasing order.
// Returns true if n has
// counts of consecutive
// 1's are increasing order.
function areSetBitsIncreasing( $n)
{
// Initialize previous count
$prev_count = PHP_INT_MAX;
// We traverse bits from right
// to left and check if counts
// are decreasing order.
while ($n > 0)
{
// Ignore 0s until we
// reach a set bit.
while ($n > 0 && $n % 2 == 0)
$n = $n / 2;
// Count current set bits
$curr_count = 1;
while ($n > 0 and $n % 2 == 1)
{
$n = $n / 2;
$curr_count++;
}
// Compare current with previous
// and update previous.
if ($curr_count >= $prev_count)
return false;
$prev_count = $curr_count;
}
return true;
}
// Driver code
$n = 10;
if (areSetBitsIncreasing($n))
echo "Yes";
else
echo "No";
// This code is contributed by anuj_67
?>
<script>
// Javascript program to check if counts of
// consecutive 1s are increasing order.
// Returns true if n has counts of
// consecutive 1's are increasing
// order.
function areSetBitsIncreasing(n)
{
// Initialize previous count
var prev_count = Number.MAX_VALUE;
// We traverse bits from right to
// left and check if counts are
// decreasing order.
while (n > 0)
{
// Ignore 0s until we reach
// a set bit.
while (n > 0 && n % 2 == 0)
n = parseInt(n / 2);
// Count current set bits
var curr_count = 1;
while (n > 0 && n % 2 == 1)
{
n = n / 2;
curr_count++;
}
// Compare current with previous
// and update previous.
if (curr_count >= prev_count)
return false;
prev_count = curr_count;
}
return true;
}
// Driver code
var n = 10;
if (areSetBitsIncreasing(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by Rajput-Ji
</script>
Time Complexity: O(log2n)
Auxiliary Space: O(1)
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