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. 

# 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

# 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

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.

# Python3 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

# Python3 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

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).

kartik

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Article Tags :

• Strings
• Pattern Searching
• Python Programs
• DSA

Practice Tags :

• Pattern Searching
• Strings