Naive algorithm for 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

Input:  txt[] =  "AABAACAADAABAABA"
        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. 

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 txt[] = "AABAACAADAABAAABAA";
    char pat[] = "AABA";
    search(pat, txt);
    return 0;
}

C++

// C++ program for Naive Pattern
// Searching algorithm
#include <bits/stdc++.h>
using namespace std;

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]
            cout << "Pattern found at index "
                 << i << endl;
    }
}

// Driver Code
int main()
{
    char txt[] = "AABAACAADAABAAABAA";
    char pat[] = "AABA";
    search(pat, txt);
    return 0;
}

// This code is contributed
// by Akanksha Rai

Java

// Java program for Naive Pattern Searching

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);
    }
}
// This code is contributed by D. Vishnu Rahul Varma

Python3

# Python3 program for Naive Pattern
# Searching algorithm
def search(pat, txt):
    M = len(pat)
    N = len(txt)

    # A loop to slide pat[] one by one */
    for i in range(N - M + 1):
        j = 0

        # For current index i, check
        # for pattern match */
        while(j < M):
            if (txt[i + j] != pat[j]):
                break
            j += 1

        if (j == M):
            print("Pattern found at index ", i)

# Driver Code
if __name__ == '__main__':
    txt = "AABAACAADAABAAABAA"
    pat = "AABA"
    search(pat, txt)

# This code is contributed
# by PrinciRaj1992

C#

// C# program for Naive Pattern Searching
using System;

class GFG {

    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[i + j] != pat[j])
                    break;

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

    // Driver code
    public static void Main()
    {
        String txt = "AABAACAADAABAAABAA";
        String pat = "AABA";
        search(txt, pat);
    }
}
// This code is Contributed by Sam007

PHP

<?php
// PHP program for Naive Pattern
// Searching algorithm

function search($pat, $txt)
{
    $M = strlen($pat);
    $N = strlen($txt);

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

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

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

    // Driver Code
    $txt = "AABAACAADAABAAABAA";
    $pat = "AABA";
    search($pat, $txt);

// This code is contributed by Sam007
?>

Javascript

<script>
    // Javascript program for Naive Pattern Searching

    function search(txt, pat)
    {
        let M = pat.length;
        let N = txt.length;

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

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

            // if pat[0...M-1] = txt[i, i+1, ...i+M-1]
            if (j == M)
                document.write("Pattern found at index " + i + "</br>");
        }
    }

    let txt = "AABAACAADAABAAABAA";
    let pat = "AABA";
    search(txt, pat);

</script>

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.

C

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. 

C

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

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

C

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

The number of comparisons in the 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. 

Alternate Approach using 2 pointers – 

In this approach we take 2 pointers i and j similar to as discussed in the above approach. In this approach instead of checking for each index again and again we move ahead as we check for our pattern string. Basically the idea behind is that once a pattern is broken there is no point to search for the pattern in the indexes that j traversed through. I know it is hard to understand but try to dry run the code and it will make sense! Code has comments to better explain each step.

This algorithm will be much better in case the pattern string is very large and there are incomplete occurrences of patterns in the main string.

eg. 

string text = “AAAAAAABBBBBBBBBBBCCCDDAAAAAAABBBBBBBBBBBCCCDDD”

string pat = “AAAAAAABBBBBBBBBBBCCCDDD”

Now here in the text at index 0, the pattern would have been found if there were 3 D’s instead of 2,  now incase we were using the approach as discussed above, we would have to again check for all indexes between 0 and index of ‘D’ but we very well know that the pattern cannot exist between these two indexes. (think about it) 

Try to dry run the code and it will make more sense!

C++

// C++ code to find patterns in a string

#include <iostream>
using namespace std;

// function that return index from where pattern begins
int search(string pat, string txt)
{
    // Your code here
    int j = 0;
    int m = pat.length();

      // notice how 'i' moves ahead with 'j' and we don't have to check for each index 'i'
    for (int i = 0; i < txt.length(); i++)
    {
        if (txt[i] == pat[j]) // trying to look for pattern
            j++;
        else
        {
            if (j > 0){ //this is very important, since incase there is no matching we will keep reducing i and get stuck in an infinite loop
                i--; //we do this because for loop will increment 'i', but we want to keep searching for the
                      //pattern from the index where the pattern was broken
            }
            j = 0; //resetting pattern
        }

        if (j == m) // if this is true then that means pattern is found!
            return i - pat.length() + 1;//since i has moved ahead with pattern
    }

    return -1; // no matching found
}

int main()
{
    string txt = "AABAACAADAABAAABAA";
    string pat =  "AABA";
      cout << "Pattern found at index : "<< search(pat,txt);
    return 0;
}

// code contributed by Anshit Bansal

Time Complexity – O(n)

Space Complexity – O(1)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.