Substring algorithm

Question

Can someone explain to me how to solve the substring problem iteratively?

The problem: given two strings S=S₁S₂S

Answer 1

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");

}

}

Answer 2

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 3

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 4

// 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 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

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;
    }