Longest Common Substring

Problem Statement

Give two string $s_1$ and $s_2$, find the longest common substring (LCS). E.g: X = [111001], Y = [11011], the longest common substring is [110] with length 3.

One terse way is to use Dynamic Programming (DP) to analyze the complex problem.

Instead of dealing with irregular substring, we can first deal with substring indexed by last character.

Define $dp[i][j] =$ the length of longest common substring of $s_1[0$~$i]$ and $s_2[0$~$j]$ ending with $s1[i]$ and $s2[j]$.

Then, the maximum LCS length could be the maximum number in array $dp$.

In order to get the value of $dp[i][j]$, we need to know if $s1[i]$ == $s2[j]$. If it is, then the $dp[i][j] = dp[i-1][j-1]+1$, else it'll be zero. Thus:

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1	dp[i][j] = (s1[i] == s2[j] ? (dp[i-1][j-1] + 1) : 0);

As we want to know the concrete string with LCM, we just need to do a few modifications.

When we get a larger $dp[i][j]$ than present maxLength, we'll update the maxLength by $dp[i][j]$.

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1 2	if(dp[i][j] > maxLen)     maxLen = dp[i][j];

At the same time, we can also record the starting index of the new longer substring. For string $s_1$, the beginning index of LCM is the present index $i$ adding 1 minus the length of LCM, i.e.

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if(dp[i][j] > maxLen){     maxLen = dp[i][j];     maxIndex = i + 1 - maxLen; }

 

Finally, we need to initialize state of $dp$. That's simple:

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for(int i = 0; i < s1.length(); ++i)     dp[i][0] = (s1[i] == s2[0] ? 1 : 0);   for(int j = 0; j < s2.length(); ++j)     dp[0][j] = (s1[0] == s2[j] ? 1 : 0);

 

 

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The complete code is:

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void LCM(const string s1, const string s2, int &sIndex, int &length) {     n1 = s1.length();     n2 = s2.length();           if(0 == n1 || 0 == n2)     {         sIndex = -1;         length = 0;         return;     }           // initialize dp     vector<vector<int> > dp;     for(int i = 0; i < n1; ++i){         vector<int> tmp;         tmp.push_back((s1[i] == s2[0] ? 1 : 0));　　// Initialize the bottom line         for(int j = 1; j < n2; ++j)         {             if(0 == i){                 tmp.push_back((s1[0] == s2[j] ? 1 : 0));　　// Initialize the left line             }else{                 tmp.push_back(0);　　// Empty the interior area             }         }                   dp.push_back(tmp);     }           // compute max length and index     length = 0;     for(int i = 1; i < n1; ++i){         for(int j = 1; j < n2; ++j){             if(st1[i] == st2[j])                 dp[i][j] = dp[i-1][j-1] + 1;                               if(dp[i][j] > length){                 length = dp[i][j];                 sIndex = i + 1 - length;             }         }     }    }