[Microsoft] Intealeaving of two given strings, Solution

Desgin an algorithm to find whether a given sting is formed by the Intealeaving of two given strings.
  For example,
stringA: ABCEF...
string B: ABCA...

dst string : ABCABCEF....

[Thoughts]
Brute-force

1:  /* a simple brute force solution */  
2:  int isInterleaving(char* a, char* b, char* c, int ai, int bi, int ci)  
3:  {  
4:   printf("entry info: %d, %d, %d,\n", ai, bi, ci);  
5:   if (a[0]=='\0' && b[0]=='\0' && c[0]=='\0')   
6:    return 1;  
7:   if (b[0]=='\0' && a[0]=='\0' && c[0]!='\0')  
8:    return 0;  
9:   int tmp1=0;   
10:   int tmp2=0;  
11:   if (a[0]!='\0' && a[0]==c[0])  
12:    tmp1 = isInterleaving(a+1, b, c+1, ai+1, bi, ci+1);  
13:   if (tmp1 == 1)  
14:    return 1;  
15:   if (b[0]!='\0' && b[0]==c[0])  
16:    tmp2 = isInterleaving(a, b+1, c+1, ai, bi+1, ci+1);  
17:   printf("res: %d %d %d\n", tmp1, tmp2, (tmp1 | tmp2));  
18:   return tmp1 | tmp2;  
19:  }  

Dynamic programming method: Consider a rectangle network of m*n, where m is the length of a and n is the length of b. Label the rows using characters in string a and cols using characters in string b. 

All that you need to do is to find a path from the top left corner to the bottom right corner. This at most requires you to examine m*n states, which gives you m*n time complexity. (Assume top left is (0, 0), then element[i][j] is a boolean which means whether there is a path from (0, 0) to (i, j) that covers the first i+j characters in the target.)

1:  int isInterleavingdp(char* a, char* b, char* str, int alen, int blen)  
2:  { int i, j;   
3:    int m[alen+1][blen+1];   
4:    memset(m, 0, (alen+1)*(blen+1)*sizeof(int));    
5:    for(i=0; i<alen; i++)  
6:    {    
7:      if (a[i]==str[i])     
8:        m[i+1][0]=1;    
9:      else   
10:        break;   
11:    }   
12:    for(j=0; j<blen; j++)  
13:    {    
14:      if (b[j] == str[j])     
15:        m[0][j+1]=1;    
16:      else     
17:        break;   
18:    }   
19:    for(i=1; i<=alen; i++)  
20:    {    
21:      for(j=1; j<=blen; j++)  
22:      {     
23:        int r1 = (m[i-1][j]>0) && a[i-1]==str[i+j-1];     
24:        int r2 = (m[i][j-1]>0) && b[j-1]==str[i+j-1];     
25:        m[i][j] = r1 | r2;    
26:      }   
27:    }   
28:    for(i=0; i<=alen; i++)  
29:    {    
30:      for(j=0; j<=blen; j++)  
31:      {     
32:        printf("%d ", m[i][j]);    
33:      }    
34:      printf("\n");   
35:    }   
36:    return m[alen][blen];  
37:  }