Occurrences of a pattern in binary representation of a number

11 Aug, 2021
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:  
Convert the binary pattern into it’s decimal representation.Take an integer all_ones, whose binary representation consists of all set bits (equal to the number of bits in the pattern).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.Now if N = pattern, it means that N contained the pattern in the end in its binary representation. So update count = count + 1.Right shift N by 1 and repeat the previous two steps until N ≥ pattern & N > 0.Print the count in the end.
Convert the binary pattern into it’s decimal representation.
Take an integer all_ones, whose binary representation consists of all set bits (equal to the number of bits in the pattern).
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.
Now if N = pattern, it means that N contained the pattern in the end in its binary representation. So update count = count + 1.
Right shift N by 1 and repeat the previous two steps until N ≥ pattern & N > 0.
Print the count in the end.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
PHP
Javascript
// 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 nint 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 codeint main(){    int n = 500;    string pat = "10";    cout << countPattern(n, pat);}
// 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 nstatic 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 codepublic static void main(String args[]){    int n = 500;    String pat = "10";    System.out.println(countPattern(n, pat));}}
# 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 ndef 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 codeif __name__ == '__main__':    n = 500    pat = "10"    print(countPattern(n, pat)) # This code is contributed by# Surendra_Gangwar
// 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 nstatic 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 codepublic static void Main(){    int n = 500;    string pat = "10";    Console.WriteLine(countPattern(n, pat));}} // This code is contributed by Ryuga
<?php// PHP program to find the number of times// pattern pat occurred in binary representation// of n. // Function to return the count of occurrence// of pat in binary representation of nfunction 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 $all_ones    for ($i = strlen($pat) - 1; $i >= 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?>
<script> // Javascript 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 nfunction countPattern( n, pat){    // To store decimal value of the pattern    let pattern_int = 0;     let 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    let all_ones = 0;     // Find values of pattern_int and all_ones    for (let i = pat.length - 1; i >= 0; i--) {        let current_bit = pat.charAt(i) - '0';        pattern_int += (power_two * current_bit);        all_ones = all_ones + power_two;        power_two = power_two * 2;    }     let 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 let n = 500;let pat = "10";document.write(countPattern(n, pat)); </script>
2
 
Time Complexity: O(N)Auxiliary Space: O(1)
SURENDRA_GANGWAR
ankthon
ihritik
jana_sayantan
pankajsharmagfg
binary-string
Algorithms
Bit Magic
C++ Programs
Pattern Searching
Searching
Strings
Searching
Strings
Bit Magic
Pattern Searching
Algorithms
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