C++ Program To Check If A String Is Substring Of Another
Last Updated :
20 Jan, 2022
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 :Â
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.Â
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.Â
C++
#include <bits/stdc++.h>
using namespace std;
int isSubstring(string s1, string s2)
{
int M = s1.length();
int N = s2.length();
for ( int i = 0; i <= N - M; i++)
{
int j;
for (j = 0; j < M; j++)
if (s2[i + j] != s1[j])
break ;
if (j == M)
return i;
}
return -1;
}
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:Â
- Time complexity: O(m * n) where m and n are lengths of s1 and s2 respectively.Â
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).Â
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:Â
Another Efficient Solution:Â
- 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++
#include <bits/stdc++.h>
using namespace std;
int Substr(string s2, string s1)
{
int counter = 0;
int i = 0;
for (; i < s1.length(); i++)
{
if (counter==s2.length())
break ;
if (s2[counter]==s1[i])
{
counter++;
}
else
{
if (counter > 0)
{
i -= counter;
}
counter = 0;
}
}
return (counter < s2.length() ?
-1 : i - counter);
}
int main()
{
string s1 =
"geeksfffffoorrfoorforgeeks" ;
cout << Substr( "for" , s1);
return 0;
}
|
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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