Occurrences of a pattern in binary representation of a number

Given a string pat and an integer N, the task is to find the number of occurrences of the pattern pat in binary representation of N.

Examples:

Input: N = 2, pat = “101”
Output: 0
Pattern “101” doesn’t occur in the binary representation of 2 (10).



Input: N = 10, pat = “101”
Output: 1
Binary representation of 10 is 1010 and the given pattern occurs only once.

Naive Approach: Convert the number into its binary string representation and then use a pattern matching algorithm to check the number of times the pattern has occurred in the binary representation.

Efficient Approach:

  1. Convert the binary pattern into it’s decimal representation.
  2. Take an integer all_ones, whose binary representation consists of all set bits (equal to the number of bits in the pattern).
  3. Performing N & all_ones now will leave only the last k bits unchanged (others will be 0) where k is the number of bits in the pattern.
  4. Now if N = pattern, it means that N contained the pattern in the end in its binary representation. So update count = count + 1.
  5. Right shift N by 1 and repeat the previous two steps until N ≥ pattern & N > 0.
  6. Print the count in the end.

Below is the implementation of the above approach:

C++



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// C++ program to find the number of times
// pattern p occurred in binary representation
// on n.
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of occurrence
// of pat in binary representation of n
int countPattern(int n, string pat)
{
    // To store decimal value of the pattern
    int pattern_int = 0;
  
    int power_two = 1;
  
    // To store a number that has all ones in
    // its binary representation and length
    // of ones equal to length of the pattern
    int all_ones = 0;
  
    // Find values of pattern_int and all_ones
    for (int i = pat.length() - 1; i >= 0; i--) {
        int current_bit = pat[i] - '0';
        pattern_int += (power_two * current_bit);
        all_ones = all_ones + power_two;
        power_two = power_two * 2;
    }
  
    int count = 0;
    while (n && n >= pattern_int) {
  
        // If the pattern occurs in the last
        // digits of n
        if ((n & all_ones) == pattern_int) {
            count++;
        }
  
        // Right shift n by 1 bit
        n = n >> 1;
    }
    return count;
}
  
// Driver code
int main()
{
    int n = 500;
    string pat = "10";
    cout << countPattern(n, pat);
}

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Java

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// Java program to find the number of times
// pattern p occurred in binary representation
// on n.
import java.util.*;
  
class solution
{
  
// Function to return the count of occurrence
// of pat in binary representation of n
static int countPattern(int n, String pat)
{
    // To store decimal value of the pattern
    int pattern_int = 0;
  
    int power_two = 1;
  
    // To store a number that has all ones in
    // its binary representation and length
    // of ones equal to length of the pattern
    int all_ones = 0;
  
    // Find values of pattern_int and all_ones
    for (int i = pat.length() - 1; i >= 0; i--) {
        int current_bit = pat.charAt(i) - '0';
        pattern_int += (power_two * current_bit);
        all_ones = all_ones + power_two;
        power_two = power_two * 2;
    }
  
    int count = 0;
    while (n!=0 && n >= pattern_int) {
  
        // If the pattern occurs in the last
        // digits of n
        if ((n & all_ones) == pattern_int) {
            count++;
        }
  
        // Right shift n by 1 bit
        n = n >> 1;
    }
    return count;
}
  
// Driver code
public static void main(String args[])
{
    int n = 500;
    String pat = "10";
    System.out.println(countPattern(n, pat));
}
}

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Python3

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# Python 3 program to find the number of times
# pattern p occurred in binary representation
# on n.
  
# Function to return the count of occurrence
# of pat in binary representation of n
def countPattern(n, pat):
      
    # To store decimal value of the pattern
    pattern_int = 0
  
    power_two = 1
  
    # To store a number that has all ones in
    # its binary representation and length
    # of ones equal to length of the pattern
    all_ones = 0
  
    # Find values of pattern_int and all_ones
    i = len(pat) - 1
    while(i >= 0):
        current_bit = ord(pat[i]) - ord('0')
        pattern_int += (power_two * current_bit)
        all_ones = all_ones + power_two
        power_two = power_two * 2
        i -= 1
      
    count = 0
    while (n != 0 and n >= pattern_int):
          
        # If the pattern occurs in the last
        # digits of n
        if ((n & all_ones) == pattern_int):
            count += 1
          
        # Right shift n by 1 bit
        n = n >> 1
      
    return count
  
# Driver code
if __name__ == '__main__':
    n = 500
    pat = "10"
    print(countPattern(n, pat))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program to find the number of times 
// pattern p occurred in binary representation 
// on n. 
using System ;
  
class GFG 
  
// Function to return the count of occurrence 
// of pat in binary representation of n 
static int countPattern(int n, string pat) 
    // To store decimal value of the pattern 
    int pattern_int = 0; 
  
    int power_two = 1; 
  
    // To store a number that has all ones in 
    // its binary representation and length 
    // of ones equal to length of the pattern 
    int all_ones = 0; 
  
    // Find values of pattern_int and all_ones 
    for (int i = pat.Length - 1; i >= 0; i--) 
    
        int current_bit = pat[i] - '0'
        pattern_int += (power_two * current_bit); 
        all_ones = all_ones + power_two; 
        power_two = power_two * 2; 
    
  
    int count = 0; 
    while (n != 0 && n >= pattern_int) 
    
  
        // If the pattern occurs in the last 
        // digits of n 
        if ((n & all_ones) == pattern_int) 
        
            count++; 
        
  
        // Right shift n by 1 bit 
        n = n >> 1; 
    
    return count; 
  
// Driver code 
public static void Main() 
    int n = 500; 
    string pat = "10"
    Console.WriteLine(countPattern(n, pat)); 
  
// This code is contributed by Ryuga

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PHP

= 0; $i–)
{
$current_bit = $pat[$i] – ‘0’;
$pattern_int += ($power_two * $current_bit);
$all_ones = $all_ones + $power_two;
$power_two = $power_two * 2;
}

$count = 0;
while ($n && $n >= $pattern_int)
{

// If the pattern occurs in the last
// digits of $n
if (($n & $all_ones) == $pattern_int)
{
$count++;
}

// Right shift $n by 1 bit
$n = $n >> 1;
}
return $count;
}

// Driver code
$n = 500;
$pat = “10”;
echo countPattern($n, $pat);

// This code is contributed by ihritik
?>

Output:

2


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