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 = "neveropen" Output: 5 Explanation: String "for" is present as a substring of s2. Input: s1 = "practice", s2 = "neveropen" Output: -1. Explanation: There is no occurrence of "practice" in "neveropen"
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
# 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 = "neveropen" 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:
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
# 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).
Please refer complete article on Check if a string is substring of another for more details!
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