Given three strings A,B and C, find if C string is interleaved of two strings. A and B. C is said to be interleaved if it contains all the characters of A and B and order of characters in respective strings is maintained. For example string C in figure is interleaved string of A and B.

# Find if string is interleaved of two strings

Simple algorithm to find if a string is interleaved of two other string is to start checking length of C as length of A + length of B. If it is not true return false (why?).

If length criteria passes, check first character of C and A. If they match, move to second character of C and A. Keep B at first character. If character in A does not match with character in C, check if first character of C and B match, if they do, then move to second character of C and B. A remains at first character.

Once done, repeat above steps with next characters of strings, while character in C matches character in A or B. If both current character in C does not match with any of the current characters in A and B, return false.

With above algorithm, with each comparison, problem reduces to smaller size, with C with length N-1, one of A or B with M-1. From the description, it is clear that recursion can be used to solve interleaving problem.

isInterleaved(A,B,C) = isInterleaved(A+1, B, C+1) || isInterleaved(A, B+1,C+1)

What shall be base case for above recursion?

If end of C is reached, all characters in C are checked, return true if all characters in other two strings are also considered. If not returned false.

Other way, if both string are A and B are fully considered and there are still characters left in C to be considered, then return false. (This case is anyways caught at length check).

### Recursive Implementation of interleaved strings algorithm

#include <stdio.h> #include <string.h> #define true 1 #define false 0; int isInterleaved(char *c, char *a, char *b){ if(!(*c) && !(*a) && !(*b)) return true; if(*c == '\0'){ return false; } // if character of a and c match return ((*c == *a) && isInterleaved(c+1,a+1,b)) || // if character of b and c match ((*c == *b) && isInterleaved(c+1,a,b+1)); } int main(void) { char *c = "abxycdz"; char *a = "abcd"; char *b = "xyz"; if(isInterleaved(c,a,b)){ printf("\n String is interleaved"); } else{ printf("\n String is not interleaved"); } return 0; }

### Iterative implementation is as follows

#include <stdio.h> #include <string.h> #define true 1 #define false 0; int isInterleaved(char *c, char *a, char *b){ while(*c != '\0'){ if(*c == *a){ a++; } else if(*c == *b){ b++; } else{ return false; } c++; } if(*a != '\0' || b != '\0') return false; return true; } int main(void) { // your code goes here char *c = "abxycdz"; char *a = "abcd"; char *b = "xyz"; if(isInterleaved(c,a,b)){ printf("\n String is interleaved"); } else{ printf("\n String is not interleaved"); } return 0; }

Iterative implementation will not work with input with common characters in string A and B, for example A = XXY and B = XXZ and if C = XXZXXY will return false.

Complexity of iterative implementation will be linear O(N), N being length of string C, where as complexity of recursive solution will b O(2^{N}) but it does not fail in when two strings have common character.

## Dynamic programming approach to interleaved strings problem

Look closely and notice that many subproblems are being calculated again and again. Recursion tree for input A = XXY and B = XXZ and C = XXZXXY

Looking at the recursion tree, it is clear that if results of subproblems are stored, they need not be calculated again and again. Create a two dimensional table. Table(i,j) = true only if C[i+j-1] is interleaved string if A[i] and B[j]. Empty string is interleaved of two other strings so, **Table[0][0] = true**

If one of the strings was empty: **Table(i,0) = A[i] == C[i] && Table(i-1, 0)** that is to say if till i-1 characters C was interleaved of A, then for ith character it will be true if i^{th} character matches i^{th} character of A. Note that B is empty. Again if string A is empty, then Table(0,j) = Table(0, j-1). With these base cases, we can fill table bottom up as follows

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

## Dynamic programming solution to find if string is interleaved of two strings

#include <stdio.h> #include <string.h> #define true 1 #define false 0; int isInterleaved(char *c, char *a, char *b){ int lenA = strlen(a); int lenB = strlen(b); int i,j; int Table[lenA+1][lenB+1]; // Initialization for(i=0; i<=lenA; i++){ for(j=0; j<=lenB; j++){ Table[i][j] = false; } } for(i=0; i<=lenA; i++){ for(j=0; j<=lenB; j++){ // Both strings are empty if(i==0 && j==0) Table[i][j] = true; // string A is empty, compare characters in C and B if(i==0 && c[j-1] == b[j-1]){ Table[i][j] = Table[i][j-1]; } // string B is empty, compare characters in C and A else if(j==0 && c[i-1] == a[i-1]){ Table[i][j] = Table[i-1][j]; } // Both strings are not empty //1. If character of A matches with character of C // but not of B else if (a[i-1] == c[i+j-1] && b[j-1] != c[i+j-1]){ Table[i][j] = Table[i-1][j]; } //2. If character of B matches with character of C // but not of A else if (a[i-1] != c[i+j-1] && b[j-1] == c[i+j-1]){ Table[i][j] = Table[i][j-1]; } //1. If character of A matches with character of C // and character of B also matches with C else if (a[i-1] == c[i+j-1] && b[j-1] == c[i+j-1]){ Table[i][j] = Table[i-1][j] || Table[i][j-1]; } } } return Table[lenA][lenB]; } int main(void) { // your code goes here char *c = "abxycdz"; char *a = "abcd"; char *b = "xyz"; if(isInterleaved(c,a,b)){ printf("\n String is interleaved"); } else{ printf("\n String is not interleaved"); } return 0; }

Complexity of dynamic programming algorithm to **find if a string is interleaved of two strings** will be O(n^{2}).

For more dynamic programming problems, please refer : Dynamic programming problems

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