Dynamic Programming | Set 33 (Find if a string is interleaved of two other strings) | GeeksforGeeks

Dynamic Programming | Set 33 (Find if a string is interleaved of two other strings) | GeeksforGeeks
Given three strings A, B and C. Write a function that checks whether C is an interleaving of A and B. C is said to be interleaving A and B, if it contains all characters of A and B and order of all characters in individual strings is preserved.

Dynamic Programming
we can solve it by creating a table and store results of subproblems in bottom up manner.
http://n00tc0d3r.blogspot.com/2013/03/interleaving-string-gene-matching.html

Construct a (n1+1)*(n2+1) table, A[0 .. n1][0 .. n2], where n1 and n2 is the length of s1 and s2, respectively.

A[i][j] = true, i.e. s3[0 .. i+j-1] is an interleaving of s1[0 .. i-1] and s2[0 .. j-1].
Thus, A[n1][n2] represents whether s3 is a interleaving of s1 and s2.

To build up such a table,

1. Set A[0][0] = true. Period.
2. Initialize the first row. The first row means whether s1[0 .. (n1-1)] matches s3[0 .. (n1-1)], correspondingly.
3. Similarly, initialize the first column which means whether s2[0 .. (n2-1)] matchess3[0 .. (n2-1)], correspondingly.
4. Now we fill up the table. For A[i][j], s3[0 .. (i+j-1)] match? s1[0 .. i-1] weaving s2[0 .. j-1]
• Note that A[i-1][j] comes from s1[0 .. i-2] weaving s2[0 .. j-1]. Thus, if it is true, we need to compare s1[i-1] with s3[i+j-1].
• Similarly, A[i][j-1] comes from s1[0 .. i-1] weaving s2[0 .. j-2]. Thus, if it is true, compare s2[j-1] with s3[i+j-1].
5. Return A[n1][n2]

public boolean isInterleave(String s1, String s2, String s3) {  
   int n1=s1.length(), n2=s2.length(), n3=s3.length();  
   if (n1 + n2 != n3) return false;  
   boolean[][] match = new boolean[n1+1][n2+1];  
   // initialize the first row and first column  
   match[0][0] = true;  
   for (int i=0; i<n1; ++i) {  
     match[i+1][0] = (s1.charAt(i) == s3.charAt(i));  
   }  
   for (int i=0; i<n2; ++i) {  
     match[0][i+1] = (s2.charAt(i) == s3.charAt(i));  
   }  
   // fill up the table  
   for (int i=0; i<n1; ++i) {  
     for (int j=0; j<n2; ++j) {  
       if ((match[i][j+1] && s3.charAt(i+j+1)==s1.charAt(i))  
           || (match[i+1][j] && s3.charAt(i+j+1)==s2.charAt(j))) {  
         match[i+1][j+1] = true;  
       }  
     }  
   }  
   return match[n1][n2];  
 }  

Space Optimization:

Notice that for each A[i][j], we only need A[i-1][j] and A[i][j-1]. So, we can improve the above algorithm with an 1D array. By doing that, the space of the new algorithm get down to O(min(n1, n2)).

 public boolean isInterleave(String s1, String s2, String s3) {  
   int n1=s1.length(), n2=s2.length(), n3=s3.length();  
   if (n1 + n2 != n3) return false;  
   // switch to save space if needed  
   if (n1<n2) {  
     String tmp = s1; s1 = s2; s2 = tmp;  
     n1 = s1.length(); n2 = s2.length();  
   }  
   boolean[] match = new boolean[n2+1];  
   // initialize the first row and first column  
   match[0] = true;  
   for (int j=0; j<n2; ++j) {  
     match[j+1] = (match[j] && s2.charAt(j)==s3.charAt(j));  
     if (!match[j+1]) break;  
   }  
   // fill up the table  
   for (int i=1; i<=n1; ++i) {  
     match[0] = (match[0] && s1.charAt(i-1)==s3.charAt(i-1));  
     for (int j=1; j<=n2; ++j) {  
       match[j] = (match[j] && s3.charAt(i+j-1)==s1.charAt(i-1))  
             || (match[j-1] && s3.charAt(i+j-1)==s2.charAt(j-1));  
     }  
   }  
   return match[n2];  
 }

Find if a string is formed by interleaving two strings - PrismoSkills
It's better to handle base cases out specially, out of loop: better performance, easier to figure out all base cases.

boolean isInterleaved(String A, String B, String C)

{

    int M = A.length();

    int N = B.length();

 

    if ((M+N) != C.length())

       return false;

 

    // lookup[i][j] is true if C[0..i+j-1] is an interleaving of A[0..i-1] and B[0..j-1].

    boolean lookup[][] = new boolean[M+1][N+1];

    for (int i=0; i<=M; i++)

    {

        for (int j=0; j<=N; j++)

        {

            // If both A and B are empty, then C must also be empty (due to length matching)

            // And since one empty string is interleaving of other two empty

            // strings, lookup[0][0] is true.

            if (i==0 && j==0)

                lookup[i][j] = true;

 

            // If A is empty, check one-to-one match with B

            else if (i==0 && B[j-1]==C[j-1])

                lookup[i][j] = lookup[i][j-1];

 

            // If B is empty, check one-to-one match with A

            else if (j==0 && A[i-1]==C[i-1])

                lookup[i][j] = lookup[i-1][j];

 

 

            // Regular check for interleaving

 

            if (A[i-1]==C[i+j-1] && B[j-1]==C[i+j-1])

                lookup[i][j]=(lookup[i-1][j] || lookup[i][j-1]);

 

            else if(A[i-1]==C[i+j-1])

                lookup[i][j] = lookup[i-1][j];

 

            else if (B[j-1]==C[i+j-1])

                lookup[i][j] = lookup[i][j-1];

 

            // Its too soon to return a false if C[i+j-1] matches neither A or B.

            // Think why?

        }

    }

 

    return lookup[M][N];

}

bool isInterleaved(char* A, char* B, char* C)

{

    // Find lengths of the two strings

    int M = strlen(A), N = strlen(B);

    // Let us create a 2D table to store solutions of

    // subproblems.  C[i][j] will be true if C[0..i+j-1]

    // is an interleaving of A[0..i-1] and B[0..j-1].

    bool IL[M+1][N+1];

    memset(IL, 0, sizeof(IL)); // Initialize all values as false.

    // C can be an interleaving of A and B only of sum

    // of lengths of A & B is equal to length of C.

    if ((M+N) != strlen(C))

       return false;

    // Process all characters of A and B

    for (int i=0; i<=M; ++i)

    {

        for (int j=0; j<=N; ++j)

        {

            // two empty strings have an empty string

            // as interleaving

            if (i==0 && j==0)

                IL[i][j] = true;

            // A is empty

            else if (i==0 && B[j-1]==C[j-1])

                IL[i][j] = IL[i][j-1];

            // B is empty

            else if (j==0 && A[i-1]==C[i-1])

                IL[i][j] = IL[i-1][j];

            // Current character of C matches with current character of A,

            // but doesn't match with current character of B

            else if(A[i-1]==C[i+j-1] && B[j-1]!=C[i+j-1])

                IL[i][j] = IL[i-1][j];

            // Current character of C matches with current character of B,

            // but doesn't match with current character of A

            else if (A[i-1]!=C[i+j-1] && B[j-1]==C[i+j-1])

                IL[i][j] = IL[i][j-1];

            // Current character of C matches with that of both A and B

            else if (A[i-1]==C[i+j-1] && B[j-1]==C[i+j-1])

                IL[i][j]=(IL[i-1][j] || IL[i][j-1]) ;

        }

    }

    return IL[M][N];

}

The worst case time complexity of recursive solution is O(2ⁿ).

a) If first character of C matches with first character of A, we move one character ahead in A and C and recursively check.

b) If first character of C matches with first character of B, we move one character ahead in B and C and recursively check.'

http://n00tc0d3r.blogspot.com/2013/03/interleaving-string-gene-matching.html

• Test a character in s1, s1[i]==s3[k]?, if succeeded, move s1 and s3 to next character, and then test the rest of s3 with the rest of s1 and s2;
• Or test a character in s2, s2[j]==s3[k]?, if succeeded, move s2 and s3 to next character, and then test the rest of s3 with the rest of s1 and s2.

a) If first character of C matches with first character of A, we move one character ahead in A and C and recursively check.

b) If first character of C matches with first character of B, we move one character ahead in B and C and recursively check.

 private boolean helper(char[] s1, int n1, char[] s2, int n2, char[] s3, int n3) {  
   if (n1==s1.length && n2==s2.length && n3==s3.length) return true;  
   if (n1==s1.length) return (s2[n2]==s3[n3] && helper(s1, n1, s2, n2+1, s3, n3+1));  
   if (n2==s2.length) return (s1[n1]==s3[n3] && helper(s1, n1+1, s2, n2, s3, n3+1));  
   return (s1[n1]==s3[n3] && helper(s1, n1+1, s2, n2, s3, n3+1))  
       || (s2[n2]==s3[n3] && helper(s1, n1, s2, n2+1, s3, n3+1));  
 }  
   
 public boolean isInterleave(String s1, String s2, String s3) {  
   if (s1.length() + s2.length() != s3.length()) return false;  
   return helper(s1.toCharArray(), 0, s2.toCharArray(), 0, s3.toCharArray(), 0);  
 } 

Cleaner:

    private static boolean isInterleaved

    (

        String s1, String s2, String s, 

        int pos1, int pos2, int pos

    ) {

        if (pos >= s.length())

            return true;

        

        boolean match1 = false;

        boolean match2 = false;

        

        if (pos1 < s1.length())

            match1 = (s.charAt(pos) == s1.charAt(pos1));

        

        if (pos2 < s2.length())

            match2 = (s.charAt(pos) == s2.charAt(pos2));

        

        return (

                (match1 && isInterleaved (s1, s2, s, pos1+1, pos2, pos+1)) ||

                (match2 && isInterleaved (s1, s2, s, pos1, pos2+1, pos+1))

            );

        

    }

bool isInterleaved(char *A, char *B, char *C)

{

    // Base Case: If all strings are empty

    if (!(*A || *B || *C))

        return true;

    // If C is empty and any of the two strings is not empty

    if (*C == '\0')

        return false;

    // If any of the above mentioned two possibilities is true,

    // then return true, otherwise false

    return ( (*C == *A) && isInterleaved(A+1, B, C+1))

           || ((*C == *B) && isInterleaved(A, B+1, C+1));

}

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