Given a text *txt[0..n-1] *and a pattern *pat[0..m-1]*, write a function *search(char pat[], char txt[])* that prints all occurrences of *pat[] *in *txt[]*. You may assume that *n > m*.

Examples:

Input: txt[] = "THIS IS A TEST TEXT" pat[] = "TEST" Output: Pattern found at index 10 Input: txt[] = "AABAACAADAABAABA" pat[] = "AABA" Output: Pattern found at index 0 Pattern found at index 9 Pattern found at index 12

## We strongly recommend that you click here and practice it, before moving on to the solution.

Pattern searching is an important problem in computer science. When we do search for a string in notepad/word file or browser or database, pattern searching algorithms are used to show the search results.

We have discussed Naive pattern searching algorithm in the previous post. The worst case complexity of Naive algorithm is O(m(n-m+1)). Time complexity of KMP algorithm is O(n) in worst case.

**KMP (Knuth Morris Pratt) Pattern Searching**

The Naive pattern searching algorithm doesn’t work well in cases where we see many matching characters followed by a mismatching character. Following are some examples.

txt[] = "AAAAAAAAAAAAAAAAAB" pat[] = "AAAAB" txt[] = "ABABABCABABABCABABABC" pat[] = "ABABAC" (not a worst case, but a bad case for Naive)

The KMP matching algorithm uses degenerating property (pattern having same sub-patterns appearing more than once in the pattern) of the pattern and improves the worst case complexity to O(n). The basic idea behind KMP’s algorithm is: whenever we detect a mismatch (after some matches), we already know some of the characters in the text of next window. We take advantage of this information to avoid matching the characters that we know will anyway match. Let us consider below example to understand this.

Matching Overviewtxt = "AAAAABAAABA" pat = "AAAA" We compare first window oftxtwithpattxt = "AAAAABAAABA" pat = "AAAA" [Initial position] We find a match. This is same as Naive String Matching. In the next step, we compare next window oftxtwithpat. txt = "AAAAABAAABA" pat = "AAAA" [Pattern shifted one position] This is where KMP does optimization over Naive. In this second window, we only compare fourth A of pattern with fourth character of current window of text to decide whether current window matches or not. Since we know first three characters will anyway match, we skipped matching first three characters.Need of Preprocessing?An important question arises from above explanation, how to know how many characters to be skipped. To know this, we pre-process pattern and prepare an integer array lps[] that tells us count of characters to be skipped.

**Preprocessing Overview:**

- KMP algorithm does preproceses pat[] and constructs an auxiliary
**lps[]**of size m (same as size of pattern) which is used to skip characters while matching. -
**name**. A proper prefix is prefix with whole string**lps**indicates longest proper prefix which is also suffix.**not**allowed. For example, prefixes of “ABC” are “”, “A”, “AB” and “ABC”. Proper prefixes are “”, “A” and “AB”. Suffixes of the string are “”, “C”, “BC” and “ABC”. - For each sub-pattern pat[0..i] where i = 0 to m-1, lps[i] stores length of the maximum matching proper prefix which is also a suffix of the sub-pattern pat[0..i].
lps[i] = the longest proper prefix of pat[0..i] which is also a suffix of pat[0..i].

**Note :** lps[i] could also be defined as longest prefix which is also proper suffix. We need to use proper at one place to make sure that the whole substring is not considered.

Examples of lps[] construction: For the pattern “AAAA”, lps[] is [0, 1, 2, 3] For the pattern “ABCDE”, lps[] is [0, 0, 0, 0, 0] For the pattern “AABAACAABAA”, lps[] is [0, 1, 0, 1, 2, 0, 1, 2, 3, 4, 5] For the pattern “AAACAAAAAC”, lps[] is [0, 1, 2, 0, 1, 2, 3, 3, 3, 4] For the pattern “AAABAAA”, lps[] is [0, 1, 2, 0, 1, 2, 3]

**Searching Algorithm:**

Unlike Naive algorithm, where we slide the pattern by one and compare all characters at each shift, we use a value from lps[] to decide the next characters to be matched. The idea is to not match character that we know will anyway match.

How to use lps[] to decide next positions (or to know number of characters to be skipped)?

- We start comparison of pat[j] with j = 0 with characters of current window of text.
- We keep matching characters txt[i] and pat[j] and keep incrementing i and j while pat[j] and txt[i] keep
**matching**. - When we see a
**mismatch**- We know that characters pat[0..j-1] match with txt[i-j+1…i-1] (Note that j starts with 0 and increment it only when there is a match).
- We also know (from above definition) that lps[j-1] is count of characters of pat[0…j-1] that are both proper prefix and suffix.
- From above two points, we can conclude that we do not need to match these lps[j-1] characters with txt[i-j…i-1] because we know that these characters will anyway match. Let us consider above example to understand this.

txt[] = "AAAAABAAABA" pat[] = "AAAA" lps[] = {0, 1, 2, 3} i = 0, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j[ match, do i++, j++ i = 1, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j[ match, do i++, j++ i = 2, j = 2 txt[] = "AAAAABAAABA" pat[] = "AAAA" pat[i] and pat[j[ match, do i++, j++ i = 3, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j[ match, do i++, j++ i = 4, j = 4 Since j == M, printpattern foundand resset j, j = lps[j-1] = lps[3] = 3 Here unlike Naive algorithm, we do not match first three characters of this window. Value of lps[j-1] (in above step) gave us index of next character to match. i = 4, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j[ match, do i++, j++ i = 5, j = 4 Since j == M, printpattern foundand reset j, j = lps[j-1] = lps[3] = 3 Again unlike Naive algorithm, we do not match first three characters of this window. Value of lps[j-1] (in above step) gave us index of next character to match. i = 5, j = 3 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[2] = 2 i = 5, j = 2 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[1] = 1 i = 5, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j > 0, change only j j = lps[j-1] = lps[0] = 0 i = 5, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] do NOT match and j is 0, we do i++. i = 6, j = 0 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++ and j++ i = 7, j = 1 txt[] = "AAAAABAAABA" pat[] = "AAAA" txt[i] and pat[j] match, do i++ and j++ We continue this way...

## C++

// C++ program for implementation of KMP pattern searching // algorithm #include<bits/stdc++.h> void computeLPSArray(char *pat, int M, int *lps); // Prints occurrences of txt[] in pat[] void KMPSearch(char *pat, char *txt) { int M = strlen(pat); int N = strlen(txt); // create lps[] that will hold the longest prefix suffix // values for pattern int lps[M]; // Preprocess the pattern (calculate lps[] array) computeLPSArray(pat, M, lps); int i = 0; // index for txt[] int j = 0; // index for pat[] while (i < N) { if (pat[j] == txt[i]) { j++; i++; } if (j == M) { printf("Found pattern at index %d \n", i-j); j = lps[j-1]; } // mismatch after j matches else if (i < N && pat[j] != txt[i]) { // Do not match lps[0..lps[j-1]] characters, // they will match anyway if (j != 0) j = lps[j-1]; else i = i+1; } } } // Fills lps[] for given patttern pat[0..M-1] void computeLPSArray(char *pat, int M, int *lps) { // length of the previous longest prefix suffix int len = 0; lps[0] = 0; // lps[0] is always 0 // the loop calculates lps[i] for i = 1 to M-1 int i = 1; while (i < M) { if (pat[i] == pat[len]) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { // This is tricky. Consider the example. // AAACAAAA and i = 7. The idea is similar // to search step. if (len != 0) { len = lps[len-1]; // Also, note that we do not increment // i here } else // if (len == 0) { lps[i] = 0; i++; } } } } // Driver program to test above function int main() { char *txt = "ABABDABACDABABCABAB"; char *pat = "ABABCABAB"; KMPSearch(pat, txt); return 0; }

## Java

// JAVA program for implementation of KMP pattern // searching algorithm class KMP_String_Matching { void KMPSearch(String pat, String txt) { int M = pat.length(); int N = txt.length(); // create lps[] that will hold the longest // prefix suffix values for pattern int lps[] = new int[M]; int j = 0; // index for pat[] // Preprocess the pattern (calculate lps[] // array) computeLPSArray(pat,M,lps); int i = 0; // index for txt[] while (i < N) { if (pat.charAt(j) == txt.charAt(i)) { j++; i++; } if (j == M) { System.out.println("Found pattern "+ "at index " + (i-j)); j = lps[j-1]; } // mismatch after j matches else if (i < N && pat.charAt(j) != txt.charAt(i)) { // Do not match lps[0..lps[j-1]] characters, // they will match anyway if (j != 0) j = lps[j-1]; else i = i+1; } } } void computeLPSArray(String pat, int M, int lps[]) { // length of the previous longest prefix suffix int len = 0; int i = 1; lps[0] = 0; // lps[0] is always 0 // the loop calculates lps[i] for i = 1 to M-1 while (i < M) { if (pat.charAt(i) == pat.charAt(len)) { len++; lps[i] = len; i++; } else // (pat[i] != pat[len]) { // This is tricky. Consider the example. // AAACAAAA and i = 7. The idea is similar // to search step. if (len != 0) { len = lps[len-1]; // Also, note that we do not increment // i here } else // if (len == 0) { lps[i] = len; i++; } } } } // Driver program to test above function public static void main(String args[]) { String txt = "ABABDABACDABABCABAB"; String pat = "ABABCABAB"; new KMP_String_Matching().KMPSearch(pat,txt); } } // This code has been contributed by Amit Khandelwal.

## Python

# Python program for KMP Algorithm def KMPSearch(pat, txt): M = len(pat) N = len(txt) # create lps[] that will hold the longest prefix suffix # values for pattern lps = [0]*M j = 0 # index for pat[] # Preprocess the pattern (calculate lps[] array) computeLPSArray(pat, M, lps) i = 0 # index for txt[] while i < N: if pat[j] == txt[i]: i += 1 j += 1 if j == M: print "Found pattern at index " + str(i-j) j = lps[j-1] # mismatch after j matches elif i < N and pat[j] != txt[i]: # Do not match lps[0..lps[j-1]] characters, # they will match anyway if j != 0: j = lps[j-1] else: i += 1 def computeLPSArray(pat, M, lps): len = 0 # length of the previous longest prefix suffix lps[0] # lps[0] is always 0 i = 1 # the loop calculates lps[i] for i = 1 to M-1 while i < M: if pat[i]==pat[len]: len += 1 lps[i] = len i += 1 else: # This is tricky. Consider the example. # AAACAAAA and i = 7. The idea is similar # to search step. if len != 0: len = lps[len-1] # Also, note that we do not increment i here else: lps[i] = 0 i += 1 txt = "ABABDABACDABABCABAB" pat = "ABABCABAB" KMPSearch(pat, txt) # This code is contributed by Bhavya Jain

Output:

Found pattern at index 10

**Preprocessing Algorithm:**

In the preprocessing part, we calculate values in lps[]. To do that, we keep track of the length of the longest prefix suffix value (we use len variable for this purpose) for the previous index. We initialize lps[0] and len as 0. If pat[len] and pat[i] match, we increment len by 1 and assign the incremented value to lps[i]. If pat[i] and pat[len] do not match and len is not 0, we update len to lps[len-1]. See computeLPSArray () in the below code for details.

**Illustration of preprocessing (or construction of lps[])**

pat[] = "AAACAAAA" len = 0, i = 0.lps[0] is always 0, we move to i = 1 len = 0, i = 1. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 1,lps[1] = 1, i = 2 len = 1, i = 2. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 2,lps[2] = 2, i = 3 len = 2, i = 3. Since pat[len] and pat[i] do not match, and len > 0, set len = lps[len-1] = lps[1] = 1 len = 1, i = 3. Since pat[len] and pat[i] do not match and len > 0, len = lps[len-1] = lps[0] = 0 len = 0, i = 3. Since pat[len] and pat[i] do not match and len = 0, Setlps[3] = 0and i = 4. len = 0, i = 4. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 1,lps[4] = 1, i = 5 len = 1, i = 5. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 2,lps[5] = 2, i = 6 len = 2, i = 6. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 3,lps[6] = 3, i = 7 len = 3, i = 7. Since pat[len] and pat[i] do not match and len > 0, set len = lps[len-1] = lps[2] = 2 len = 2, i = 7. Since pat[len] and pat[i] match, do len++, store it in lps[i] and do i++. len = 3,lps[7] = 3, i = 8 We stop here as we have constructed the whole lps[].

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