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

• Difficulty Level : Easy
• Last Updated : 12 Jun, 2022

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 using namespace std; // Returns true if n has increasing count of// continuous-1 else falsebool findContinuous1(int n){    const int bits = 8 * sizeof(int);     // store the bit-pattern of n into    // bit bitset- bp    string bp = bitset(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 codeint main(){    int n = 179;    if (findContinuous1(n))        cout << "Yes";    else        cout << "No";     return 0;}

## Java

 // 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

 # 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 coden = 179if (findContinuous1(n)):    print("Yes")else:    print("No") # This code is contributed by SHUBHAMSINGH10

## C#

 // 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

## Javascript



Output :

Yes

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.#includeusing 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 codeint main(){    int n = 10;    if (areSetBitsIncreasing(n))        cout << "Yes";    else        cout << "No";     return 0;}

## Java

 // 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

 # 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 coden = 10 if (areSetBitsIncreasing(n)):    print("Yes")else:    print("No")     # This code is contributed by Smitha

## C#

 // 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

 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?>

## Javascript



Output :

No

Time Complexity: O(log2n)

Auxiliary Space: O(1)