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Boyer Moore Algorithm

Itchy Impala answered on November 16, 2022 Popularity 10/10 Helpfulness 1/10

answer Boyer Moore Algorithm

Boyer Moore Algorithm

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/* Java program for Boyer Moore Algorithm with
Good Suffix heuristic to find pattern in
given text string */
class GFG
{

// preprocessing for strong good suffix rule
static void preprocess_strong_suffix(int []shift, int []bpos,
									char []pat, int m)
{
	// m is the length of pattern
	int i = m, j = m + 1;
	bpos[i] = j;

	while(i > 0)
	{
		/*if character at position i-1 is not
		equivalent to character at j-1, then
		continue searching to right of the
		pattern for border */
		while(j <= m && pat[i - 1] != pat[j - 1])
		{
			/* the character preceding the occurrence of t
			in pattern P is different than the mismatching
			character in P, we stop skipping the occurrences
			and shift the pattern from i to j */
			if (shift[j] == 0)
				shift[j] = j - i;

			//Update the position of next border
			j = bpos[j];
		}
		/* p[i-1] matched with p[j-1], border is found.
		store the beginning position of border */
		i--; j--;
		bpos[i] = j;
	}
}

//Preprocessing for case 2
static void preprocess_case2(int []shift, int []bpos,
							char []pat, int m)
{
	int i, j;
	j = bpos[0];
	for(i = 0; i <= m; i++)
	{
		/* set the border position of the first character
		of the pattern to all indices in array shift
		having shift[i] = 0 */
		if(shift[i] == 0)
			shift[i] = j;

		/* suffix becomes shorter than bpos[0],
		use the position of next widest border
		as value of j */
		if (i == j)
			j = bpos[j];
	}
}

/*Search for a pattern in given text using
Boyer Moore algorithm with Good suffix rule */
static void search(char []text, char []pat)
{
	// s is shift of the pattern
	// with respect to text
	int s = 0, j;
	int m = pat.length;
	int n = text.length;

	int []bpos = new int[m + 1];
	int []shift = new int[m + 1];

	//initialize all occurrence of shift to 0
	for(int i = 0; i < m + 1; i++)
		shift[i] = 0;

	//do preprocessing
	preprocess_strong_suffix(shift, bpos, pat, m);
	preprocess_case2(shift, bpos, pat, m);

	while(s <= n - m)
	{
		j = m - 1;

		/* Keep reducing index j of pattern while
		characters of pattern and text are matching
		at this shift s*/
		while(j >= 0 && pat[j] == text[s+j])
			j--;

		/* If the pattern is present at the current shift,
		then index j will become -1 after the above loop */
		if (j < 0)
		{
			System.out.printf("pattern occurs at shift = %d\n", s);
			s += shift[0];
		}
		else
		
			/*pat[i] != pat[s+j] so shift the pattern
			shift[j+1] times */
			s += shift[j + 1];
	}

}

// Driver Code
public static void main(String[] args)
{
	char []text = "ABAAAABAACD".toCharArray();
	char []pat = "ABA".toCharArray();
	search(text, pat);
}
}

Popularity 10/10 Helpfulness 1/10 Language java

Source: favtutor.com

Tags: algorithm boyer-moore java

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Contributed on Nov 16 2022

Itchy Impala

0 Answers Avg Quality 2/10

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