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.
Below is the implementation of the above approach:
// 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);
} |
// 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));
} } |
# 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 |
// 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 |
<?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 n function 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 n function 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
Complexity Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)