Given two strings X and Y, find the Longest Common Substring of X and Y.

Naive [O(N*M^{2})] and Dynamic Programming [O(N*M)] approaches are already discussed here.

In this article, we will discuss a linear time approach to find LCS using suffix tree (The 5^{th} Suffix Tree Application).

Here we will build generalized suffix tree for two strings X and Y as discussed already at:

Generalized Suffix Tree 1

Lets take same example (X = xabxa, and Y = babxba) we saw in Generalized Suffix Tree 1.

We built following suffix tree for X and Y there:

This is generalized suffix tree for xabxa#babxba$

In above, leaves with suffix indices in [0,4] are suffixes of string xabxa and leaves with suffix indices in [6,11] are suffixes of string babxa. Why ??

Because in concatenated string xabxa#babxba$, index of string xabxa is 0 and it’s length is 5, so indices of it’s suffixes would be 0, 1, 2, 3 and 4. Similarly index of string babxba is 6 and it’s length is 6, so indices of it’s suffixes would be 6, 7, 8, 9, 10 and 11.

With this, we can see that in the generalized suffix tree figure above, there are some internal nodes having leaves below it from

- both strings X and Y (i.e. there is at least one leaf with suffix index in [0,4] and one leaf with suffix index in [6, 11]
- string X only (i.e. all leaf nodes have suffix indices in [0,4])
- string Y only (i.e. all leaf nodes have suffix indices in [6,11])

Following figure shows the internal nodes marked as “XY”, “X” or “Y” depending on which string the leaves belong to, that they have below themselves.

What these “XY”, “X” or “Y” marking mean ?

Path label from root to an internal node gives a substring of X or Y or both.

For node marked as XY, substring from root to that node belongs to both strings X and Y.

For node marked as X, substring from root to that node belongs to string X only.

For node marked as Y, substring from root to that node belongs to string Y only.

By looking at above figure, can you see how to get LCS of X and Y ?

By now, it should be clear that how to get common substring of X and Y at least.

If we traverse the path from root to nodes marked as XY, we will get common substring of X and Y.

Now we need to find the longest one among all those common substrings.

Can you think how to get LCS now ? Recall how did we get Longest Repeated Substring in a given string using suffix tree already.

The path label from root to the deepest node marked as XY will give the LCS of X and Y. The deepest node is highlighted in above figure and path label “abx” from root to that node is the LCS of X and Y.

// A C program to implement Ukkonen's Suffix Tree Construction // Here we build generalized suffix tree for two strings // And then we find longest common substring of the two input strings #include <stdio.h> #include <string.h> #include <stdlib.h> #define MAX_CHAR 256 struct SuffixTreeNode { struct SuffixTreeNode *children[MAX_CHAR]; //pointer to other node via suffix link struct SuffixTreeNode *suffixLink; /*(start, end) interval specifies the edge, by which the node is connected to its parent node. Each edge will connect two nodes, one parent and one child, and (start, end) interval of a given edge will be stored in the child node. Lets say there are two nods A and B connected by an edge with indices (5, 8) then this indices (5, 8) will be stored in node B. */ int start; int *end; /*for leaf nodes, it stores the index of suffix for the path from root to leaf*/ int suffixIndex; }; typedef struct SuffixTreeNode Node; char text[100]; //Input string Node *root = NULL; //Pointer to root node /*lastNewNode will point to newly created internal node, waiting for it's suffix link to be set, which might get a new suffix link (other than root) in next extension of same phase. lastNewNode will be set to NULL when last newly created internal node (if there is any) got it's suffix link reset to new internal node created in next extension of same phase. */ Node *lastNewNode = NULL; Node *activeNode = NULL; /*activeEdge is represeted as input string character index (not the character itself)*/ int activeEdge = -1; int activeLength = 0; // remainingSuffixCount tells how many suffixes yet to // be added in tree int remainingSuffixCount = 0; int leafEnd = -1; int *rootEnd = NULL; int *splitEnd = NULL; int size = -1; //Length of input string int size1 = 0; //Size of 1st string Node *newNode(int start, int *end) { Node *node =(Node*) malloc(sizeof(Node)); int i; for (i = 0; i < MAX_CHAR; i++) node->children[i] = NULL; /*For root node, suffixLink will be set to NULL For internal nodes, suffixLink will be set to root by default in current extension and may change in next extension*/ node->suffixLink = root; node->start = start; node->end = end; /*suffixIndex will be set to -1 by default and actual suffix index will be set later for leaves at the end of all phases*/ node->suffixIndex = -1; return node; } int edgeLength(Node *n) { if(n == root) return 0; return *(n->end) - (n->start) + 1; } int walkDown(Node *currNode) { /*activePoint change for walk down (APCFWD) using Skip/Count Trick (Trick 1). If activeLength is greater than current edge length, set next internal node as activeNode and adjust activeEdge and activeLength accordingly to represent same activePoint*/ if (activeLength >= edgeLength(currNode)) { activeEdge += edgeLength(currNode); activeLength -= edgeLength(currNode); activeNode = currNode; return 1; } return 0; } void extendSuffixTree(int pos) { /*Extension Rule 1, this takes care of extending all leaves created so far in tree*/ leafEnd = pos; /*Increment remainingSuffixCount indicating that a new suffix added to the list of suffixes yet to be added in tree*/ remainingSuffixCount++; /*set lastNewNode to NULL while starting a new phase, indicating there is no internal node waiting for it's suffix link reset in current phase*/ lastNewNode = NULL; //Add all suffixes (yet to be added) one by one in tree while(remainingSuffixCount > 0) { if (activeLength == 0) activeEdge = pos; //APCFALZ // There is no outgoing edge starting with // activeEdge from activeNode if (activeNode->children1] == NULL) { //Extension Rule 2 (A new leaf edge gets created) activeNode->children1] = newNode(pos, &leafEnd); /*A new leaf edge is created in above line starting from an existng node (the current activeNode), and if there is any internal node waiting for it's suffix link get reset, point the suffix link from that last internal node to current activeNode. Then set lastNewNode to NULL indicating no more node waiting for suffix link reset.*/ if (lastNewNode != NULL) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } } // There is an outgoing edge starting with activeEdge // from activeNode else { // Get the next node at the end of edge starting // with activeEdge Node *next = activeNode->children1]; if (walkDown(next))//Do walkdown { //Start from next node (the new activeNode) continue; } /*Extension Rule 3 (current character being processed is already on the edge)*/ if (text[next->start + activeLength] == text[pos]) { //If a newly created node waiting for it's //suffix link to be set, then set suffix link //of that waiting node to curent active node if(lastNewNode != NULL && activeNode != root) { lastNewNode->suffixLink = activeNode; lastNewNode = NULL; } //APCFER3 activeLength++; /*STOP all further processing in this phase and move on to next phase*/ break; } /*We will be here when activePoint is in middle of the edge being traversed and current character being processed is not on the edge (we fall off the tree). In this case, we add a new internal node and a new leaf edge going out of that new node. This is Extension Rule 2, where a new leaf edge and a new internal node get created*/ splitEnd = (int*) malloc(sizeof(int)); *splitEnd = next->start + activeLength - 1; //New internal node Node *split = newNode(next->start, splitEnd); activeNode->children1] = split; //New leaf coming out of new internal node split->children1] = newNode(pos, &leafEnd); next->start += activeLength; split->children1] = next; /*We got a new internal node here. If there is any internal node created in last extensions of same phase which is still waiting for it's suffix link reset, do it now.*/ if (lastNewNode != NULL) { /*suffixLink of lastNewNode points to current newly created internal node*/ lastNewNode->suffixLink = split; } /*Make the current newly created internal node waiting for it's suffix link reset (which is pointing to root at present). If we come across any other internal node (existing or newly created) in next extension of same phase, when a new leaf edge gets added (i.e. when Extension Rule 2 applies is any of the next extension of same phase) at that point, suffixLink of this node will point to that internal node.*/ lastNewNode = split; } /* One suffix got added in tree, decrement the count of suffixes yet to be added.*/ remainingSuffixCount--; if (activeNode == root && activeLength > 0) //APCFER2C1 { activeLength--; activeEdge = pos - remainingSuffixCount + 1; } else if (activeNode != root) //APCFER2C2 { activeNode = activeNode->suffixLink; } } } void print(int i, int j) { int k; for (k=i; k<=j && text[k] != '#'; k++) printf("%c", text[k]); if(k<=j) printf("#"); } //Print the suffix tree as well along with setting suffix index //So tree will be printed in DFS manner //Each edge along with it's suffix index will be printed void setSuffixIndexByDFS(Node *n, int labelHeight) { if (n == NULL) return; if (n->start != -1) //A non-root node { //Print the label on edge from parent to current node //Uncomment below line to print suffix tree //print(n->start, *(n->end)); } int leaf = 1; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { //Uncomment below two lines to print suffix index // if (leaf == 1 && n->start != -1) // printf(" [%d]\n", n->suffixIndex); //Current node is not a leaf as it has outgoing //edges from it. leaf = 0; setSuffixIndexByDFS(n->children[i], labelHeight + edgeLength(n->children[i])); } } if (leaf == 1) { for(i= n->start; i<= *(n->end); i++) { if(text[i] == '#') { n->end = (int*) malloc(sizeof(int)); *(n->end) = i; } } n->suffixIndex = size - labelHeight; //Uncomment below line to print suffix index // printf(" [%d]\n", n->suffixIndex); } } void freeSuffixTreeByPostOrder(Node *n) { if (n == NULL) return; int i; for (i = 0; i < MAX_CHAR; i++) { if (n->children[i] != NULL) { freeSuffixTreeByPostOrder(n->children[i]); } } if (n->suffixIndex == -1) free(n->end); free(n); } /*Build the suffix tree and print the edge labels along with suffixIndex. suffixIndex for leaf edges will be >= 0 and for non-leaf edges will be -1*/ void buildSuffixTree() { size = strlen(text); int i; rootEnd = (int*) malloc(sizeof(int)); *rootEnd = - 1; /*Root is a special node with start and end indices as -1, as it has no parent from where an edge comes to root*/ root = newNode(-1, rootEnd); activeNode = root; //First activeNode will be root for (i=0; i<size; i++) extendSuffixTree(i); int labelHeight = 0; setSuffixIndexByDFS(root, labelHeight); } int doTraversal(Node *n, int labelHeight, int* maxHeight, int* substringStartIndex) { if(n == NULL) { return; } int i=0; int ret = -1; if(n->suffixIndex < 0) //If it is internal node { for (i = 0; i < MAX_CHAR; i++) { if(n->children[i] != NULL) { ret = doTraversal(n->children[i], labelHeight + edgeLength(n->children[i]), maxHeight, substringStartIndex); if(n->suffixIndex == -1) n->suffixIndex = ret; else if((n->suffixIndex == -2 && ret == -3) || (n->suffixIndex == -3 && ret == -2) || n->suffixIndex == -4) { n->suffixIndex = -4;//Mark node as XY //Keep track of deepest node if(*maxHeight < labelHeight) { *maxHeight = labelHeight; *substringStartIndex = *(n->end) - labelHeight + 1; } } } } } else if(n->suffixIndex > -1 && n->suffixIndex < size1)//suffix of X return -2;//Mark node as X else if(n->suffixIndex >= size1)//suffix of Y return -3;//Mark node as Y return n->suffixIndex; } void getLongestCommonSubstring() { int maxHeight = 0; int substringStartIndex = 0; doTraversal(root, 0, &maxHeight, &substringStartIndex); int k; for (k=0; k<maxHeight; k++) printf("%c", text[k + substringStartIndex]); if(k == 0) printf("No common substring"); else printf(", of length: %d",maxHeight); printf("\n"); } // driver program to test above functions int main(int argc, char *argv[]) { size1 = 7; printf("Longest Common Substring in xabxac and abcabxabcd is: "); strcpy(text, "xabxac#abcabxabcd$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); size1 = 10; printf("Longest Common Substring in xabxaabxa and babxba is: "); strcpy(text, "xabxaabxa#babxba$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); size1 = 14; printf("Longest Common Substring in GeeksforGeeks and GeeksQuiz is: "); strcpy(text, "GeeksforGeeks#GeeksQuiz$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); size1 = 26; printf("Longest Common Substring in OldSite:GeeksforGeeks.org"); printf(" and NewSite:GeeksQuiz.com is: "); strcpy(text, "OldSite:GeeksforGeeks.org#NewSite:GeeksQuiz.com$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); size1 = 6; printf("Longest Common Substring in abcde and fghie is: "); strcpy(text, "abcde#fghie$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); size1 = 6; printf("Longest Common Substring in pqrst and uvwxyz is: "); strcpy(text, "pqrst#uvwxyz$"); buildSuffixTree(); getLongestCommonSubstring(); //Free the dynamically allocated memory freeSuffixTreeByPostOrder(root); return 0; }

Output:

Longest Common Substring in xabxac and abcabxabcd is: abxa, of length: 4 Longest Common Substring in xabxaabxa and babxba is: abx, of length: 3 Longest Common Substring in GeeksforGeeks and GeeksQuiz is: Geeks, of length: 5 Longest Common Substring in OldSite:GeeksforGeeks.org and NewSite:GeeksQuiz.com is: Site:Geeks, of length: 10 Longest Common Substring in abcde and fghie is: e, of length: 1 Longest Common Substring in pqrst and uvwxyz is: No common substring

If two strings are of size M and N, then Generalized Suffix Tree construction takes O(M+N) and LCS finding is a DFS on tree which is again O(M+N).

So overall complexity is linear in time and space.

**Followup:**

- Given a pattern, check if it is substring of X or Y or both. If it is a substring, find all it’s occurrences along with which string (X or Y or both) it belongs to.
- Extend the implementation to find LCS of more than two strings
- Solve problem 1 for more than two strings
- Given a string, find it’s Longest Palindromic Substring

We have published following more articles on suffix tree applications:

- Suffix Tree Application 1 – Substring Check
- Suffix Tree Application 2 – Searching All Patterns
- Suffix Tree Application 3 – Longest Repeated Substring
- Suffix Tree Application 4 – Build Linear Time Suffix Array
- Suffix Tree Application 6 – Longest Palindromic Substring
- Generalized Suffix Tree 1

This article is contributed by **Anurag Singh**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above