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++
#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;
}
|
Java
class GFG {
static 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.charAt(i + j)
!= s1.charAt(j))
break;
if (j == M)
return i;
}
return -1;
}
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);
}
}
|
Python3
def isSubstring(s1, s2):
M = len(s1)
N = len(s2)
for i in range(N - M + 1):
for j in range(M):
if (s2[i + j] != s1[j]):
break
if j + 1 == M :
return i
return -1
if __name__ == "__main__":
s1 = "for"
s2 = "geeksforgeeks"
res = isSubstring(s1, s2)
if res == -1 :
print("Not present")
else:
print("Present at index " + str(res))
|
C#
using System;
class GFG {
static 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;
}
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);
}
}
|
PHP
<?php
function isSubstring($s1, $s2)
{
$M = strlen($s1);
$N = strlen($s2);
for ($i = 0; $i <= $N - $M; $i++)
{
$j = 0;
for (; $j < $M; $j++)
if ($s2[$i + $j] != $s1[$j])
break;
if ($j == $M)
return $i;
}
return -1;
}
$s1 = "for";
$s2 = "geeksforgeeks";
$res = isSubstring($s1, $s2);
if ($res == -1)
echo "Not present";
else
echo "Present at index " . $res;
?>
|
Javascript
<script>
function isSubstring(s1, s2)
{
var M = s1.length;
var N = s2.length;
for (var i = 0; i <= N - M; i++) {
var j;
for (j = 0; j < M; j++)
if (s2[i + j] != s1[j])
break;
if (j == M)
return i;
}
return -1;
}
var s1 = "for";
var s2 = "geeksforgeeks";
var res = isSubstring(s1, s2);
if (res == -1)
document.write( "Not present");
else
document.write( "Present at index " + res);
</script>
|
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.
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;
}
|
Java
import java.io.*;
class GFG {
public static int Substr(String s2, String s1){
int counter = 0;
int i = 0;
for(;i<s1.length();i++){
if(counter==s2.length())
break;
if(s2.charAt(counter)==s1.charAt(i)){
counter++;
}else{
if(counter>0){
i -= counter;
}
counter = 0;
}
}
return counter < s2.length()?-1:i-counter;
}
public static void main (String[] args) {
String s1 = "geeksfffffoorrfoorforgeeks";
System.out.println(Substr("for", s1));
}
}
|
Python3
def Substr(Str, target):
t = 0
Len = len(Str)
i = 0
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)
print(Substr("GeeksForGeeks", "Fr"))
print(Substr("GeeksForGeeks", "For"))
|
C#
using System;
class GFG {
static 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;
}
public static int Main()
{
string s1 = "geeksfffffoorrfoorforgeeks";
Console.Write(Substr("for", s1));
return 0;
}
}
|
Javascript
<!-- Javascript program for the above approach -->
<script>
function Substr( s2, s1){
var counter = 0;
var 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;
}
var s1 = "geeksfffffoorrfoorforgeeks";
document.write(Substr("for", s1));
</script>
<!-- this code is contributed by Nirajgusain5 -->
|
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).
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