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Check if a string is substring of another

  • Difficulty Level : Medium
  • Last Updated : 24 Sep, 2021

Given two strings s1 and s2, find if s1 is a substring of s2. If yes, return the index of the first occurrence, else return -1.

Examples : 

Input: s1 = "for", s2 = "geeksforgeeks"
Output: 5
Explanation:
String "for" is present as a substring
of s2.

Input: s1 = "practice", s2 = "geeksforgeeks"
Output: -1.
Explanation:
There is no occurrence of "practice" in
"geeksforgeeks"

Simple Approach: The idea is to run a loop from start to end and for every index in the given string check whether the sub-string can be formed from that index. This can be done by running a nested loop traversing the given string and in that loop run another loop checking for sub-string from every index. 
For example, consider there to be a string of length N and a substring of length M. Then run a nested loop, where the outer loop runs from 0 to (N-M) and the inner loop from 0 to M. For very index check if the sub-string traversed by the inner loop is the given sub-string or not. 

C++




// C++ program to check if a string is
// substring of other.
#include <bits/stdc++.h>
using namespace std;
 
// Returns true if s1 is substring of s2
int isSubstring(string s1, string s2)
{
    int M = s1.length();
    int N = s2.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 (s2[i + j] != s1[j])
                break;
 
        if (j == M)
            return i;
    }
 
    return -1;
}
 
/* Driver code */
int main()
{
    string s1 = "for";
    string s2 = "geeksforgeeks";
    int res = isSubstring(s1, s2);
    if (res == -1)
        cout << "Not present";
    else
        cout << "Present at index " << res;
    return 0;
}


Java




// Java program to check if a string is
// substring of other.
class GFG {
 
    // Returns true if s1 is substring of s2
    static int isSubstring(
        String s1, String s2)
    {
        int M = s1.length();
        int N = s2.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 (s2.charAt(i + j)
                    != s1.charAt(j))
                    break;
 
            if (j == M)
                return i;
        }
 
        return -1;
    }
 
    /* Driver code */
    public static void main(String args[])
    {
        String s1 = "for";
        String s2 = "geeksforgeeks";
 
        int res = isSubstring(s1, s2);
 
        if (res == -1)
            System.out.println("Not present");
        else
            System.out.println(
                "Present at index "
                + res);
    }
}
 
// This code is contributed by JaideepPyne.


Python3




# Python3 program to check if
# a string is substring of other.
 
# Returns true if s1 is substring of s2
def isSubstring(s1, s2):
    M = len(s1)
    N = len(s2)
 
    # A loop to slide pat[] one by one
    for i in range(N - M + 1):
 
        # For current index i,
        # check for pattern match
        for j in range(M):
            if (s2[i + j] != s1[j]):
                break
             
        if j + 1 == M :
            return i
 
    return -1
 
# Driver Code
if __name__ == "__main__":
    s1 = "for"
    s2 = "geeksforgeeks"
    res = isSubstring(s1, s2)
    if res == -1 :
        print("Not present")
    else:
        print("Present at index " + str(res))
 
# This code is contributed by ChitraNayal


C#




// C# program to check if a string is
// substring of other.
using System;
class GFG {
 
    // Returns true if s1 is substring of s2
    static int isSubstring(string s1, string s2)
    {
        int M = s1.Length;
        int N = s2.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 (s2[i + j] != s1[j])
                    break;
 
            if (j == M)
                return i;
        }
 
        return -1;
    }
 
    /* Driver code */
    public static void Main()
    {
        string s1 = "for";
        string s2 = "geeksforgeeks";
 
        int res = isSubstring(s1, s2);
 
        if (res == -1)
            Console.Write("Not present");
        else
            Console.Write("Present at index "
                          + res);
    }
}
 
// This code is contributed by nitin mittal.


PHP




<?php
// PHP program to check if a
// string is substring of other.
 
// Returns true if s1 is substring of s2
function isSubstring($s1, $s2)
{
    $M = strlen($s1);
    $N = strlen($s2);
 
    // A loop to slide
    // pat[] one by one
    for ($i = 0; $i <= $N - $M; $i++)
    {
        $j = 0;
 
        // For current index i,
        // check for pattern match
        for (; $j < $M; $j++)
            if ($s2[$i + $j] != $s1[$j])
                break;
 
        if ($j == $M)
            return $i;
    }
 
    return -1;
}
 
// Driver Code
$s1 = "for";
$s2 = "geeksforgeeks";
$res = isSubstring($s1, $s2);
if ($res == -1)
    echo "Not present";
else
    echo "Present at index " . $res;
 
// This code is contributed by mits
?>


Javascript




<script>
 
// JavaScript program to check if a string is
// substring of other.
 
// Returns true if s1 is substring of s2
function isSubstring(s1, s2)
{
    var M = s1.length;
    var N = s2.length;
 
    /* A loop to slide pat[] one by one */
    for (var i = 0; i <= N - M; i++) {
        var j;
 
        /* For current index i, check for
 pattern match */
        for (j = 0; j < M; j++)
            if (s2[i + j] != s1[j])
                break;
 
        if (j == M)
            return i;
    }
 
    return -1;
}
 
/* Driver code */
var s1 = "for";
var s2 = "geeksforgeeks";
var res = isSubstring(s1, s2);
if (res == -1)
    document.write( "Not present");
else
    document.write( "Present at index " + res);
 
</script>


Output

Present at index 5

Complexity Analysis: 

  • Time complexity: O(m * n) where m and n are lengths of s1 and s2 respectively. 
    A nested loop is used the outer loop runs from 0 to N-M and inner loop from 0 to M so the complexity is O(m*n).
  • Space Complexity: O(1). 
    As no extra space is required.

An efficient solution is to use a O(n) searching algorithm like KMP algorithm, Z algorithm, etc.
Language implementations: 

Another Efficient Solution: 

  • An efficient solution would need only one traversal i.e. O(n) on the longer string s1. Here we will start traversing the string s1 and maintain a pointer for string s2 from 0th index.
  • For each iteration we compare the current character in s1 and check it with the pointer at s2.
  • If they match we increment the pointer on s2 by 1. And for every mismatch we set the pointer back to 0.
  • Also keep a check when the s2 pointer value is equal to the length of string s2, if true we break and return the value (pointer of string s1 – pointer of string s2)
  • Works with strings containing duplicate characters.

C++




#include <bits/stdc++.h>
using namespace std;
 
int Substr(string s2, string s1)
{
    int counter = 0; // pointing s2
    int i = 0;
    for(;i<s1.length();i++)
    {
        if(counter==s2.length())
            break;
        if(s2[counter]==s1[i])
        {
            counter++;
        }
      else
        {
            // Special case where character preceding the i'th character is duplicate
            if(counter > 0)
            {
                i -= counter;
            }
            counter = 0;
        }
    }
    return counter < s2.length()?-1:i-counter;
}
 
// Driver code
int main()
{
    string s1 = "geeksfffffoorrfoorforgeeks";
    cout << Substr("for", s1);
    return 0;
}
 
// this code is contributed by Manu Pathria


Java




import java.io.*;
 
class GFG {
   public static int Substr(String s2, String s1){
        int counter = 0; //pointing s2
        int i = 0;
        for(;i<s1.length();i++){
            if(counter==s2.length())
                break;
            if(s2.charAt(counter)==s1.charAt(i)){
                counter++;
            }else{
                //Special case where character preceding the i'th character is duplicate
                if(counter>0){
                    i -= counter;
                }
                counter = 0;
            }
        }
        return counter < s2.length()?-1:i-counter;
    }
    public static void main (String[] args) {
        String s1 = "geeksfffffoorrfoorforgeeks";
        //System.out.println(s2.indexOf("for"));
        System.out.println(Substr("for", s1));
    }
}


Python3




# Python program for the above approach
def Substr(Str, target):
     
    t = 0
    Len = len(Str)
    i = 0
     
    # Iterate from 0 to Len - 1
    for i in range(Len):
        if (t == len(target)):
            break
        if (Str[i] == target[t]):
            t += 1
        else:
            t = 0
             
    if (t < len(target)):
        return -1
    else:
        return (i - t)
 
# Driver code
print(Substr("GeeksForGeeks", "Fr"))
print(Substr("GeeksForGeeks", "For"))
 
# This code is contributed by avanitrachhadiya2155


Javascript




<!-- Javascript program for the above approach -->
 
<script>
   function Substr( s2,  s1){
    
    var counter = 0; // pointing s2
    var i = 0;
    for( ;i < s1.length; i++)
    {
        if( counter == s2.length )
        {
            break;
        }
        if( s2[counter] == s1[i] )
        {
            counter++;
        }
      else
        {
            // Special case where character preceding
            // the i'th character is duplicate
            if(counter > 0)
            {
                i -= counter;
            }
            counter = 0;
        }
    }
    
    return counter < s2.length ? -1 : i-counter;
}
  
 
// Driver code
 var s1 = "geeksfffffoorrfoorforgeeks";
 document.write(Substr("for", s1));
  
</script>
 
<!-- this code is contributed by Nirajgusain5 -->


Output

18

Complexity Analysis:

The complexity of the above code will be still O(n*m) in the worst case and the space complexity is O(1).


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