Javascript Program To Check If A String Is Substring Of Another

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. 

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.
Please refer complete article on Check if a string is substring of another for more details!