# Searching for Patterns | Set 1 (Naive Pattern Searching)

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

pat[] =  "AABA"
Output: Pattern found at index 0
Pattern found at index 9
Pattern found at index 12

```

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.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Naive Pattern Searching:
Slide the pattern over text one by one and check for a match. If a match is found, then slides by 1 again to check for subsequent matches.

## C

```// C program for Naive Pattern Searching algorithm
#include <stdio.h>
#include <string.h>

void search(char* pat, char* txt)
{
int M = strlen(pat);
int N = strlen(txt);

/* A loop to slide pat[] one by one */
for (int i = 0; i <= N - M; i++) {
int j;

/* For current index i, check for pattern match */
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break;

if (j == M) // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
printf("Pattern found at index %d n", i);
}
}

/* Driver program to test above function */
int main()
{
char pat[] = "AABA";
search(pat, txt);
return 0;
}
```

## Python

```# Python program for Naive Pattern Searching
def search(pat, txt):
M = len(pat)
N = len(txt)

# A loop to slide pat[] one by one
for i in xrange(N-M + 1):

# For current index i, check for pattern match
for j in xrange(M):
if txt[i + j] != pat[j]:
break

if j == M-1: # if pat[0...M-1] = txt[i, i + 1, ...i + M-1]
print "Pattern found at index " + str(i)

# Driver program to test the above function
pat = "AABA"
search (pat, txt)

# This code is contributed by Bhavya Jain
```

## Java

```// Java program for Naive Pattern Searching
public class NaiveSearch {

public static void search(String txt, String pat)
{
int M = pat.length();
int N = txt.length();

/* A loop to slide pat one by one */
for (int i = 0; i <= N - M; i++) {

int j;

/* For current index i, check for pattern
match */
for (j = 0; j < M; j++)
if (txt.charAt(i + j) != pat.charAt(j))
break;

if (j == M) // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
System.out.println("Pattern found at index " + i);
}
}

public static void main(String[] args)
{
String pat = "AABA";
search(txt, pat);
}
}
// This code is contributed by Harikishore
```

Output:
```Pattern found at index 0
Pattern found at index 9
Pattern found at index 13 ```

What is the best case?
The best case occurs when the first character of the pattern is not present in text at all.

```txt[] = "AABCCAADDEE";
pat[] = "FAA";```

The number of comparisons in best case is O(n).

What is the worst case ?
The worst case of Naive Pattern Searching occurs in following scenarios.
1) When all characters of the text and pattern are same.

```txt[] = "AAAAAAAAAAAAAAAAAA";
pat[] = "AAAAA";```

2) Worst case also occurs when only the last character is different.

```txt[] = "AAAAAAAAAAAAAAAAAB";
pat[] = "AAAAB";```

Number of comparisons in worst case is O(m*(n-m+1)). Although strings which have repeated characters are not likely to appear in English text, they may well occur in other applications (for example, in binary texts). The KMP matching algorithm improves the worst case to O(n). We will be covering KMP in the next post. Also, we will be writing more posts to cover all pattern searching algorithms and data structures.

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