Recursive function to do substring search

Given a text txt[] and a pattern pat[], write a recursive function “contains(char pat[], char txt[])” that returns true if pat[] is present in txt[], otherwise false. Examples:

1) Input:   txt[] =  "THIS IS A TEST TEXT"
            pat[] = "TEST"
  Output:  true


2) Input:  txt[] =  "geeksforgeeks"
           pat[] = "quiz"
  Output:  false;

We strongly recommend to minimize the browser and try this yourself first. Below is recursive algorithm.

contains(tex[], pat[])
    1) If the current character is the last character of the text, but pat
       has more characters, return false.

    2) Else If the current character is the last character of the pattern,
       then return true

    3) Else If current characters of pat and text match, then
        return contains(text + 1, pat + 1);

    4) Else If current characters of pat and text don't match
        return contains(text + 1, pat);

Below is the implementation of the above algorithm. 

C++

// Recursive C++ program to find if a given pattern is
// present in a text
#include<iostream>
using namespace std;

bool exactMatch(char *text, char *pat)
{
 if (*text == '\0' && *pat != '\0')
  return false;

 // Else If last character of pattern reaches
 if (*pat == '\0')
  return true;

 if (*text == *pat)
  return exactMatch(text + 1, pat + 1);

 return false;
}

// This function returns true if 'text' contain 'pat'
bool contains(char *text, char *pat)
{
 // If last character of text reaches
 if (*text == '\0')
  return false;

 // If current characters of pat and text match
 if (*text == *pat)
  if(exactMatch(text, pat))
   return 1;
  else
  return contains(text + 1, pat);

 // If current characters of pat and tex don't match
 return contains(text + 1, pat);
}

// Driver program to test above function
int main()
{
 cout << contains("geeksforgeeks", "geeks") << endl;
 cout << contains("geeksforgeeks", "geeksquiz") << endl;
 cout << contains("geeksquizgeeks", "quiz") << endl;
 return 0;
}

Java

// Recursive Java program to find if a given pattern is
// present in a text

class GFG {
    static int exactMatch(String text, String pat, int text_index, int pat_index) {
        if (text_index == text.length() && pat_index != pat.length())
            return 0;

        // Else If last character of pattern reaches
        if (pat_index == pat.length())
            return 1;

        if (text.charAt(text_index) == pat.charAt(pat_index))
            return exactMatch(text, pat, text_index + 1, pat_index + 1);

        return 0;
    }

    // This function returns true if 'text' contain 'pat'
    static int contains(String text, String pat, int text_index, int pat_index) {
        // If last character of text reaches
        if (text_index == text.length())
            return 0;

        // If current characters of pat and text match
        if (text.charAt(text_index) == pat.charAt(pat_index)) {
            if (exactMatch(text, pat, text_index, pat_index) == 1)
                return 1;
            else
                return contains(text, pat, text_index + 1, pat_index);
        }

        // If current characters of pat and tex don't match
        return contains(text, pat, text_index + 1, pat_index);
    }

    // Driver program to test the above function
    public static void main(String args[]) {
        System.out.println(contains("geeksforgeeks", "geeks", 0, 0));
        System.out.println(contains("geeksforgeeks", "geeksquiz", 0, 0));
        System.out.println(contains("geeksquizgeeks", "quiz", 0, 0));

    }
}
// This code is contributed by Saurabh Jaiswal

Python3

# Recursive Python3 program to find if a given pattern is
# present in a text

def exactMatch(text, pat, text_index, pat_index):
    if text_index == len(text) and pat_index != len(pat):
        return 0

    # Else If last character of pattern reaches
    if pat_index == len(pat):
        return 1

    if text[text_index] == pat[pat_index]:
        return exactMatch(text, pat, text_index+1, pat_index+1)

    return 0


# This function returns true if 'text' contain 'pat'
def contains(text, pat, text_index, pat_index):
    # If last character of text reaches
    if text_index == len(text):
        return 0

    # If current characters of pat and text match
    if text[text_index] == pat[pat_index]:
        if exactMatch(text, pat, text_index, pat_index):
            return 1
        else:
            return contains(text, pat, text_index+1, pat_index)

    # If current characters of pat and tex don't match
    return contains(text , pat, text_index+1, pat_index)

# Driver program to test the above function

print(contains("geeksforgeeks", "geeks", 0, 0))
print(contains("geeksforgeeks", "geeksquiz", 0, 0))
print(contains("geeksquizgeeks", "quiz", 0, 0))

# This code is contributed by ankush_953.

Javascript

<script>

// Recursive JavaScript program to find if a given pattern is
// present in a text

function exactMatch(text, pat, text_index, pat_index){
    if(text_index == text.length && pat_index != pat.length)
        return 0

    // Else If last character of pattern reaches
    if(pat_index == pat.length)
        return 1

    if(text[text_index] == pat[pat_index])
        return exactMatch(text, pat, text_index+1, pat_index+1)

    return 0
}


// This function returns true if 'text' contain 'pat'
function contains(text, pat, text_index, pat_index){
    // If last character of text reaches
    if(text_index == text.length)
        return 0

    // If current characters of pat and text match
    if(text[text_index] == pat[pat_index]){
        if(exactMatch(text, pat, text_index, pat_index))
            return 1
        else
            return contains(text, pat, text_index+1, pat_index)
    }

    // If current characters of pat and tex don't match
    return contains(text , pat, text_index+1, pat_index)
}

// Driver program to test the above function

document.write(contains("geeksforgeeks", "geeks", 0, 0),"</br>")
document.write(contains("geeksforgeeks", "geeksquiz", 0, 0),"</br>")
document.write(contains("geeksquizgeeks", "quiz", 0, 0),"</br>")

// This code is contributed by shinjanpatra.

</script>

Output:

1
0
1

Time Complexity : O(n*m)
Auxiliary Space : O(n+m)

This article is contributed by Bhupinder. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Python3

def str_search(st, text, k):
    n = len(st)
    m = len(text)
    if k < n-m+1:
        if st[k:k+m] == text:
            return True
        else:
            return str_search(st, text, k+1)

    else:
        return False


t = 'this is a test text'
print(str_search(t, 'test', 0))

Output

True