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++ 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;
}

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: 

• Java Substring
• substr in C++
• Python find

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++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

int Substr(string s2, string s1)
{
	// pointing s2
	int counter = 0;
	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

Output:

18

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