Check if bits of a number has count of consecutive set bits in increasing order

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Output :

Yes

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right


Output :

No

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m, Smitha Dinesh Semwal



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.