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++
// CPP 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 program to test above function */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 program to test above function */ 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. |
Python 3
# Python 3 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 program to test above function */ 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 ?> |
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:
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