String Matching Algorithms


String matching is one of the most studied issues in Computer Science. There are three well known algorithms by

It is very unlikely that a developer in this day will be asked to write one let develop a new one. However the reason
for a refresher is that some problems may be modeled as a string matching problem. Besides, this is the actual reason
for the refresher,
they are interesting.

Classic String Match

The classic problem is described as: Given a string Y and a search string x is there a substring of Y that matches exactly with x. In C syntax a brute force algorithm is shown below

int  strSearch(const char *Y, int N, const char *x, int n)
// N is the length of string Y
// n is the length of string x
  for(int i=0; i+n <= N; ++i)
    bool found=true;
    for(j=0; found && j<n; ++j)
      found = (Y[i+j] == x[j]);
    if (found)
       return i;
 return -1;


An interesting variation is to find a substring in Y that is a permutation of the string x. Let’s try test driven development: Given a solution let us write code to verify it. In other words for some k such that 0<= k < N-n, verify that the substring Y[k, k+n) is a a permutation of X. There are n! (! denotes factorial) permutations of the substring Y[k,k+n). Comparing every permutation is not viable. One solution is to reduce both strings to a canonical form that can then be compared. One such form is to sort the two strings which makes comparison possible. Thus to find a permuted substring, just slide a window of size n across Y and check if it is a permuted version of x.

int  strPermuteSearch(const char *Y, int N, const char *x, int n)
// N is the length of string Y
// n is the length of string x
   char y[n];
   for(int i=0;  i+n <= N; ++i)
     strncpy(y, Y+i, n);
     if (strncmp(x,y,n)==0)
     return i;
   return -1;

The time complexity of this algorithm is O(N*n*Log(n)). The n*Log(n) is because of the sorting withing the loop. This can reduced to O(N*n) by using insertion sort. Since y is sorted all that is needed when i is incremented is to remove Y[i-1] from y and insert Y[i+n-1] which can be done in O(n) while keeping y sorted.


This leads to another problem. Is there a substring of Y such that some combination of n characters of the substring is a permutation of x. Again consider a verification algorithm. Given a string Z of length M where M>=n is there some collection n characters which can be permuted to match x. There are C(M,n) combinations. Trying all the combinations is not viable. Consider Z[0]. If Z[0] is present in x, our solution space is now reduced to
C(M-1,n-1). If Z[0] is not present then our solution space is reduced to C(M-1,n). The algorithm terminates when n==0 which means that a result has been found or n>M which means that Z does not contain a substring which can be permuted to x.

bool Verify(const char* Z, int M, char *x, int n)
 if (n==0) return true;
  if (n>M) return false;
  int i=0;
  for(; i<n; ++i)
   if(*Z == x[i]) break;
  if (i<n)    {
//swap x[i] with x[n-1]
char  t = x[i]; x[i]=x[n-1]; x[n-1] =t;
return Verify(Z+1, M-1, x, n-1);
}    else
return Verify(Z+1,M-1,x,n);

Notice that the linear search in Lines 5 and 6 push the time complexity to O(n*M). If the two strings were sorted, then we could reduce it to O(M) as shown below

bool VerifySorted(const char* Z, int M, char *x, int n) {
if (n==0) return true;
else if (n>M) return false;
   else if (*Z < *x)
return Verify(Z+1, M-1,x,n);
else if (*Z == *x)
return Verify(Z+1,M-1,x+1,n-1);
else // *Z>*x
      return false;

This solution can be adapted to checking if B is a subset of A, if A and B are sequences. Now let us restate the problem adding a simple restriction: Find the smallest substring of Y such that there exists n characters in it, which can be permuted
to match x. This can be done by sliding a window of size M where n<=M<=N and calling Verify with that window as shown below

struct position
  int start;
  int length;
position CombinationSubset(const char* Y, int N, char *x, int n){
  position result = {-1,-1};
  char Z[N];
  for(M=n; M<=N; ++M)  {
    for(i=0; i+M <=N; ++i)    {
     if (VerifySorted(Z+i, M, x,n))     {
       result.length = M;
       return result;
  return result;

Notice that we could make some optimisation by checking that Z[i] and Z[i+M-1] are both in x before calling VerifySorted. As before, sort(Z,Z+M) in lines 11 and 12 can be reduced to an insertion taking O(M) time.

Distinct characters

Problem:List all sub strings of length n in a string Y such that every character in the substring is unique.
Hint 1: Find the number of duplicate characters in every substring of length n.
Hint 2: use C++ sort and unique and find the difference in length

Problem: List all substrings with exactly one duplicate.
Hint: Modify the previous solution.


About The Sunday Programmer

Joe is an experienced C++/C# developer on Windows. Currently looking out for an opening in C/C++ on Windows or Linux.
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