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.
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 = "neveropen"; 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!
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