Substring algorithm - string

Question

Can someone explain to me how to solve the substring problem iteratively?
The problem: given two strings S=S1S2S3…Sn and T=T1T2T3…Tm, with m is less than or equal to n, determine if T is a substring of S.

Answer 1

Here's a list of string searching algorithms
Depending on your needs, a different algorithm may be a better fit, but Boyer-Moore is a popular choice.

Answer 2

A naive algorithm would be to test at each position 0 < i ≤ n-m of S if Si+1Si+2…Si+m=T1T2…Tm. For n=7 and m=5:
i=0: S1S2S3S4S5S6S7
| | | | |
T1T2T3T4T5
i=1: S1S2S3S4S5S6S7
| | | | |
T1T2T3T4T5
i=2: S1S2S3S4S5S6S7
| | | | |
T1T2T3T4T5
The algorithm in pseudo-code:
// we just need to test if n ≤ m
IF n > m:
// for each offset on that T can start to be substring of S
FOR i FROM 0 TO n-m:
// compare every character of T with the corresponding character in S plus the offset
FOR j FROM 1 TO m:
// if characters are equal
IF S[i+j] == T[j]:
// if we’re at the end of T, T is a substring of S
IF j == m:
RETURN true;
ENDIF;
ELSE:
BREAK;
ENDIF;
ENDFOR;
ENDFOR;
ENDIF;
RETURN false;

Answer 3

Not sure what language you're working in, but here's an example in C#. It's a roughly n2 algorithm, but it will get the job done.
bool IsSubstring (string s, string t)
{
for (int i = 0; i <= (s.Length - t.Length); i++)
{
bool found = true;
for (int j = 0; found && j < t.Length; j++)
{
if (s[i + j] != t[j])
found = false;
}
if (found)
return true;
}
return false;
}

Answer 4

if (T == string.Empty) return true;
for (int i = 0; i <= S.Length - T.Length; i++) {
for (int j = 0; j < T.Length; j++) {
if (S[i + j] == T[j]) {
if (j == (T.Length - 1)) return true;
}
else break;
}
}
return false;

Answer 5

It would go something like this:
m==0? return true
cs=0
ct=0
loop
cs>n-m? break
char at cs+ct in S==char at ct in T?
yes:
ct=ct+1
ct==m? return true
no:
ct=0
cs=cs+1
end loop
return false

Answer 6

This may be redundant with the above list of substring algorithms, but I was always amused by KMP (http://en.wikipedia.org/wiki/Knuth–Morris–Pratt_algorithm)

Answer 7

// runs in best case O(n) where no match, worst case O(n2) where strings match
var s = "hippopotumus"
var t = "tum"
for(var i=0;i<s.length;i++)
if(s[i]==t[0])
for(var ii=i,iii=0; iii<t.length && i<s.length; ii++, iii++){
if(s[ii]!=t[iii]) break
else if (iii==t.length-1) console.log("yay found it at index: "+i)
}

Answer 8

Here is my PHP variation that includes a check to make sure the Needle does not exceed the Haystacks length during the search.
<?php
function substring($haystack,$needle) {
if("" == $needle) { return true; }
echo "Haystack:\n$haystack\n";
echo "Needle:\n$needle\n";
for($i=0,$len=strlen($haystack);$i<$len;$i++){
if($needle[0] == $haystack[$i]) {
$found = true;
for($j=0,$slen=strlen($needle);$j<$slen;$j++) {
if($j >= $len) { return false; }
if($needle[$j] != $haystack[$i+$j]) {
$found = false;
continue;
}
}
if($found) {
echo " . . . . . . SUCCESS!!!! startPos: $i\n";
return true;
}
}
}
echo " . . . . . . FAILURE!\n" ;
return false;
}
assert(substring("haystack","hay"));
assert(!substring("ack","hoy"));
assert(substring("hayhayhay","hayhay"));
assert(substring("mucho22","22"));
assert(!substring("str","string"));
?>
Left in some echo's. Remove if they offend you!

Answer 9

Is a O(n*m) algorithm, where n and m are the size of each string.
In C# it would be something similar to:
public static bool IsSubtring(char[] strBigger, char[] strSmall)
{
int startBigger = 0;
while (startBigger <= strBigger.Length - strSmall.Length)
{
int i = startBigger, j = 0;
while (j < strSmall.Length && strSmall[j] == strBigger[i])
{
i++;
j++;
}
if (j == strSmall.Length)
return true;
startBigger++;
}
return false;
}

Answer 10

I know I'm late to the game but here is my version of it (in C#):
bool isSubString(string subString, string supraString)
{
for (int x = 0; x <= supraString.Length; x++)
{
int counter = 0;
if (subString[0] == supraString[x]) //find initial match
{
for (int y = 0; y <= subString.Length; y++)
{
if (subString[y] == supraString[y+x])
{
counter++;
if (counter == subString.Length)
{
return true;
}
}
}
}
}
return false;
}

Answer 11

Though its pretty old post, I am trying to answer it. Kindly correct me if anything is wrong,
package com.amaze.substring;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class CheckSubstring {
/**
* #param args
* #throws IOException
*/
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Please enter the main string");
String mainStr = br.readLine();
System.out.println("Enter the substring that has to be searched");
String subStr = br.readLine();
char[] mainArr = new char[mainStr.length()];
mainArr = mainStr.toCharArray();
char[] subArr = new char[subStr.length()];
subArr = subStr.toCharArray();
boolean tracing = false;
//System.out.println("Length of substring is "+subArr.length);
int j = 0;
for(int i=0; i<mainStr.length();i++){
if(!tracing){
if(mainArr[i] == subArr[j]){
tracing = true;
j++;
}
} else {
if (mainArr[i] == subArr[j]){
//System.out.println(mainArr[i]);
//System.out.println(subArr[j]);
j++;
System.out.println("Value of j is "+j);
if((j == subArr.length)){
System.out.println("SubString found");
return;
}
} else {
j=0;
tracing = false;
}
}
}
System.out.println("Substring not found");
}
}