Substring algorithm

Question

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

The problem: given two strings S=S₁S₂S

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

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 3

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 4

A naive algorithm would be to test at each position 0 < i ≤ n-m of S if S_i+1S_i+2…S_i+m=T₁T₂…T_m. 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 5

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;
}