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.
Input: N = 2, pat = “101”
Pattern “101” doesn’t occur in the binary representation of 2 (10).
Input: N = 10, pat = “101”
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.
- 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:
= 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)
// Right shift $n by 1 bit
$n = $n >> 1;
// Driver code
$n = 500;
$pat = “10”;
echo countPattern($n, $pat);
// This code is contributed by ihritik
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