Given text string with length n and a pattern with length m, the task is to prints all occurrences of pattern in text.
Note: You may assume that n > m.
Examples:
Input: text = “THIS IS A TEST TEXT”, pattern = “TEST”
Output: Pattern found at index 10
Input: text = “AABAACAADAABAABA”, pattern = “AABA”
Output: Pattern found at index 0, Pattern found at index 9, Pattern found at index 12

Pattern searching
Naive Pattern Searching algorithm:
Slide the pattern over text one by one and check for a match. If a match is found, then slide by 1 again to check for subsequent matches.
C++
#include <bits/stdc++.h>
using namespace std;
void search(string& pat, string txt)
{
int M = pat.size();
int N = txt.size();
for ( int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j
== M)
cout << "Pattern found at index " << i << endl;
}
}
int main()
{
string txt = "AABAACAADAABAAABAA" ;
string pat = "AABA" ;
search(pat, txt);
return 0;
}
|
C
#include <stdio.h>
#include <string.h>
void search( char * pat, char * txt)
{
int M = strlen (pat);
int N = strlen (txt);
for ( int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j
== M)
printf ( "Pattern found at index %d \n" , i);
}
}
int main()
{
char txt[] = "AABAACAADAABAAABAA" ;
char pat[] = "AABA" ;
search(pat, txt);
return 0;
}
|
Java
public class NaiveSearch {
static void search(String pat, String txt)
{
int l1 = pat.length();
int l2 = txt.length();
int i = 0 , j = l2 - 1 ;
for (i = 0 , j = l2 - 1 ; j < l1;) {
if (txt.equals(pat.substring(i, j + 1 ))) {
System.out.println( "Pattern found at index "
+ i);
}
i++;
j++;
}
}
public static void main(String args[])
{
String pat = "AABAACAADAABAAABAA" ;
String txt = "AABA" ;
search(pat, txt);
}
}
|
Python3
def search(pat, txt):
M = len (pat)
N = len (txt)
for i in range (N - M + 1 ):
j = 0
while (j < M):
if (txt[i + j] ! = pat[j]):
break
j + = 1
if (j = = M):
print ( "Pattern found at index " , i)
if __name__ = = '__main__' :
txt = "AABAACAADAABAAABAA"
pat = "AABA"
search(pat, txt)
|
C#
using System;
class GFG {
public static void search(String txt, String pat)
{
int M = pat.Length;
int N = txt.Length;
for ( int i = 0; i <= N - M; i++) {
int j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j == M)
Console.WriteLine( "Pattern found at index "
+ i);
}
}
public static void Main()
{
String txt = "AABAACAADAABAAABAA" ;
String pat = "AABA" ;
search(txt, pat);
}
}
|
Javascript
function search(txt, pat)
{
let M = pat.length;
let N = txt.length;
for (let i = 0; i <= N - M; i++) {
let j;
for (j = 0; j < M; j++)
if (txt[i + j] != pat[j])
break ;
if (j == M)
document.write( "Pattern found at index " + i + "</br>" );
}
}
let txt = "AABAACAADAABAAABAA" ;
let pat = "AABA" ;
search(txt, pat);
|
PHP
<?php
function search( $pat , $txt )
{
$M = strlen ( $pat );
$N = strlen ( $txt );
for ( $i = 0; $i <= $N - $M ; $i ++)
{
for ( $j = 0; $j < $M ; $j ++)
if ( $txt [ $i + $j ] != $pat [ $j ])
break ;
if ( $j == $M )
echo "Pattern found at index " , $i . "\n" ;
}
}
$txt = "AABAACAADAABAAABAA" ;
$pat = "AABA" ;
search( $pat , $txt );
?>
|
Output
Pattern found at index 0
Pattern found at index 9
Pattern found at index 13
Time Complexity: O(N2)
Auxiliary Space: O(1)
Complexity Analysis of Naive algorithm for Pattern Searching:
Best Case: O(n)
- When the pattern is found at the very beginning of the text (or very early on).
- The algorithm will perform a constant number of comparisons, typically on the order of O(n) comparisons, where n is the length of the pattern.
Worst Case: O(n2)
- When the pattern doesn’t appear in the text at all or appears only at the very end.
- The algorithm will perform O((n-m+1)*m) comparisons, where n is the length of the text and m is the length of the pattern.
- In the worst case, for each position in the text, the algorithm may need to compare the entire pattern against the text.
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Last Updated :
20 Oct, 2023
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