Longest Common Extension / LCE using RMQ
Prerequisites :
The Longest Common Extension (LCE) problem considers a string s and computes, for each pair (L , R), the longest sub string of s that starts at both L and R. In LCE, in each of the query we have to answer the length of the longest common prefix starting at indexes L and R.
Example:
String : āabbababbaā
Queries: LCE(1, 2), LCE(1, 6) and LCE(0, 5)Find the length of the Longest Common Prefix starting at index given as, (1, 2), (1, 6) and (0, 5).
The string highlighted āgreenā are the longest common prefix starting at index- L and R of the respective queries. We have to find the length of the longest common prefix starting at index- (1, 2), (1, 6) and (0, 5).
In Set 1, we explained about the naive method to find the length of the LCE of a string on many queries. In this set we will show how a LCE problem can be reduced to a RMQ problem, hence decreasing the asymptotic time complexity of the naive method.
Reduction of LCE to RMQ
Let the input string be S and queries be of the formLCE(L, R). Let the suffix array for s be Suff[] and the lcp array be lcp[].
The longest common extension between two suffixes SL and SR of S can be obtained from the lcp array in the following way.
- Let low be the rank of SL among the suffixes of S (that is, Suff[low] = L).
- Let high be the rank of SR among the suffixes of S. Without loss of generality, we assume that low < high.
- Then the longest common extension of SL and SR is lcp(low, high) = min (low<=k< high)lcp [k].
Proof: Let SL = SLā¦SL+Cā¦sn and SR = SRā¦SR+cā¦sn, and let c be the longest common extension of SL and SR(i.e. SLā¦SL+C-1 = snā¦SR+c-1). We assume that the string S has a sentinel character so that no suffix of S is a prefix of any other suffix of S but itself.
- If low = high ā 1 then i = low and lcp[low] = c is the longest common extension of SL and SR and we are done.
- If low < high -1 then select i such lcp[i] is the minimum value in the interval [low, high] of the lcp array. We then have two possible cases:
- If c < lcp[i] we have a contradiction because SL . . . SL+lcp[i]-1 = SR. . . SR+lcp[i]-1 by the definition of the LCP table, and the fact that the entries of lcp correspond to sorted suffixes of S.
- if c > lcp[i], let high = Suff[i], so that Shigh is the suffix associated with position i. Si is such that shigh . . . shigh+lcp[i]-1 = SL . . . SL+lcp[i]-1 and shigh . . . shigh+lcp[i]-1 = SR . . . SR+lcp[i]-1, but since SL . . . SL+c-1 = SR. . . SR+c-1 we have that the lcp array should be wrongly sorted which is a contradiction.
Therefore we have c = lcp[i]
Thus we have reduced our longest common extension query to a range minimum-query over a range in lcp.
Algorithm
- To find low and high, we must have to compute the suffix array first and then from the suffix array we compute the inverse suffix array.
- We also need lcp array, hence we use Kasaiās Algorithm to find lcp array from the suffix array.
- Once the above things are done, we simply find the minimum value in lcp array from index ā low to high (as proved above) for each query.
The minimum value is the length of the LCE for that query.
Implementation
C++
// A C++ Program to find the length of longest common // extension using Direct Minimum Algorithm #include <bits/stdc++.h> using namespace std; // Structure to represent a query of form (L,R) struct Query { int L, R; }; // Structure to store information of a suffix struct suffix { int index; // To store original index int rank[2]; // To store ranks and next rank pair }; // A utility function to get minimum of two numbers int minVal( int x, int y) { return (x < y) ? x : y; } // A utility function to get minimum of two numbers int maxVal( int x, int y) { return (x > y) ? x : y; } // A comparison function used by sort() to compare // two suffixes Compares two pairs, returns 1 if // first pair is smaller int cmp( struct suffix a, struct suffix b) { return (a.rank[0] == b.rank[0]) ? (a.rank[1] < b.rank[1]) : (a.rank[0] < b.rank[0]); } // This is the main function that takes a string 'txt' // of size n as an argument, builds and return the // suffix array for the given string vector< int > buildSuffixArray(string txt, int n) { // A structure to store suffixes and their indexes struct suffix suffixes[n]; // Store suffixes and their indexes in an array // of structures. // The structure is needed to sort the suffixes // alphabetically and maintain their old indexes // while sorting for ( int i = 0; i < n; i++) { suffixes[i].index = i; suffixes[i].rank[0] = txt[i] - 'a' ; suffixes[i].rank[1] = ((i + 1) < n) ? (txt[i + 1] - 'a' ) : -1; } // Sort the suffixes using the comparison function // defined above. sort(suffixes, suffixes + n, cmp); // At his point, all suffixes are sorted according // to first 2 characters. Let us sort suffixes // according to first 4/ characters, then first 8 // and so on // This array is needed to get the index in suffixes[] // from original index. This mapping is needed to get // next suffix. int ind[n]; for ( int k = 4; k < 2 * n; k = k * 2) { // Assigning rank and index values to first suffix int rank = 0; int prev_rank = suffixes[0].rank[0]; suffixes[0].rank[0] = rank; ind[suffixes[0].index] = 0; // Assigning rank to suffixes for ( int i = 1; i < n; i++) { // If first rank and next ranks are same as // that of previous/ suffix in array, assign // the same new rank to this suffix if (suffixes[i].rank[0] == prev_rank && suffixes[i].rank[1] == suffixes[i - 1].rank[1]) { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = rank; } else // Otherwise increment rank and assign { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = ++rank; } ind[suffixes[i].index] = i; } // Assign next rank to every suffix for ( int i = 0; i < n; i++) { int nextindex = suffixes[i].index + k / 2; suffixes[i].rank[1] = (nextindex < n) ? suffixes[ind[nextindex]].rank[0] : -1; } // Sort the suffixes according to first k characters sort(suffixes, suffixes + n, cmp); } // Store indexes of all sorted suffixes in the suffix // array vector< int > suffixArr; for ( int i = 0; i < n; i++) suffixArr.push_back(suffixes[i].index); // Return the suffix array return suffixArr; } /* To construct and return LCP */ vector< int > kasai(string txt, vector< int > suffixArr, vector< int >& invSuff) { int n = suffixArr.size(); // To store LCP array vector< int > lcp(n, 0); // Fill values in invSuff[] for ( int i = 0; i < n; i++) invSuff[suffixArr[i]] = i; // Initialize length of previous LCP int k = 0; // Process all suffixes one by one starting from // first suffix in txt[] for ( int i = 0; i < n; i++) { /* If the current suffix is at n-1, then we donāt have next substring to consider. So lcp is not defined for this substring, we put zero. */ if (invSuff[i] == n - 1) { k = 0; continue ; } /* j contains index of the next substring to be considered to compare with the present substring, i.e., next string in suffix array */ int j = suffixArr[invSuff[i] + 1]; // Directly start matching from k'th index as // at-least k-1 characters will match while (i + k < n && j + k < n && txt[i + k] == txt[j + k]) k++; lcp[invSuff[i]] = k; // lcp for the present suffix. // Deleting the starting character from the string. if (k > 0) k--; } // return the constructed lcp array return lcp; } // A utility function to find longest common extension // from index - L and index - R int LCE(vector< int > lcp, vector< int > invSuff, int n, int L, int R) { // Handle the corner case if (L == R) return (n - L); int low = minVal(invSuff[L], invSuff[R]); int high = maxVal(invSuff[L], invSuff[R]); int length = lcp[low]; for ( int i = low + 1; i < high; i++) { if (lcp[i] < length) length = lcp[i]; } return (length); } // A function to answer queries of longest common extension void LCEQueries(string str, int n, Query q[], int m) { // Build a suffix array vector< int > suffixArr = buildSuffixArray(str, str.length()); // An auxiliary array to store inverse of suffix array // elements. For example if suffixArr[0] is 5, the // invSuff[5] would store 0. This is used to get next // suffix string from suffix array. vector< int > invSuff(n, 0); // Build a lcp vector vector< int > lcp = kasai(str, suffixArr, invSuff); for ( int i = 0; i < m; i++) { int L = q[i].L; int R = q[i].R; printf ( "LCE (%d, %d) = %d\n" , L, R, LCE(lcp, invSuff, n, L, R)); } return ; } // Driver Program to test above functions int main() { string str = "abbababba" ; int n = str.length(); // LCA Queries to answer Query q[] = { { 1, 2 }, { 1, 6 }, { 0, 5 } }; int m = sizeof (q) / sizeof (q[0]); LCEQueries(str, n, q, m); return (0); } |
Java
import java.util.Arrays; // Structure to represent a query of form (L, R) class Query { int L, R; public Query( int L, int R) { this .L = L; this .R = R; } } // Structure to store information of a suffix class Suffix implements Comparable<Suffix> { int index; int [] rank = new int [ 2 ]; public Suffix( int index) { this .index = index; } @Override public int compareTo(Suffix other) { // Compare two suffixes based on their ranks return ( this .rank[ 0 ] == other.rank[ 0 ]) ? Integer.compare( this .rank[ 1 ], other.rank[ 1 ]) : Integer.compare( this .rank[ 0 ], other.rank[ 0 ]); } } public class LongestCommonExtension { // A utility function to get the minimum of two numbers static int minVal( int x, int y) { return (x < y) ? x : y; } // A utility function to get the maximum of two numbers static int maxVal( int x, int y) { return (x > y) ? x : y; } // A comparison function used by Arrays.sort() to // compare two suffixes Compares two pairs, returns 1 if // the first pair is smaller static int compare(Suffix a, Suffix b) { return (a.rank[ 0 ] == b.rank[ 0 ]) ? Integer.compare(a.rank[ 1 ], b.rank[ 1 ]) : Integer.compare(a.rank[ 0 ], b.rank[ 0 ]); } // This is the main function that takes a string 'txt' // of size n as an argument, builds and returns the // suffix array for the given string static int [] buildSuffixArray(String txt) { int n = txt.length(); Suffix[] suffixes = new Suffix[n]; for ( int i = 0 ; i < n; i++) { suffixes[i] = new Suffix(i); suffixes[i].rank[ 0 ] = txt.charAt(i) - 'a' ; suffixes[i].rank[ 1 ] = (i + 1 ) < n ? txt.charAt(i + 1 ) - 'a' : - 1 ; } // Sort the suffixes using the comparison function Arrays.sort(suffixes); int [] ind = new int [n]; for ( int k = 4 ; k < 2 * n; k = k * 2 ) { int rank = 0 ; int prev_rank = suffixes[ 0 ].rank[ 0 ]; suffixes[ 0 ].rank[ 0 ] = rank; ind[suffixes[ 0 ].index] = 0 ; for ( int i = 1 ; i < n; i++) { if (suffixes[i].rank[ 0 ] == prev_rank && suffixes[i].rank[ 1 ] == suffixes[i - 1 ].rank[ 1 ]) { prev_rank = suffixes[i].rank[ 0 ]; suffixes[i].rank[ 0 ] = rank; } else { prev_rank = suffixes[i].rank[ 0 ]; suffixes[i].rank[ 0 ] = ++rank; } ind[suffixes[i].index] = i; } for ( int i = 0 ; i < n; i++) { int nextindex = suffixes[i].index + k / 2 ; suffixes[i].rank[ 1 ] = (nextindex < n) ? suffixes[ind[nextindex]].rank[ 0 ] : - 1 ; } // Sort the suffixes according to the first k // characters Arrays.sort(suffixes); } int [] suffixArr = new int [n]; for ( int i = 0 ; i < n; i++) { suffixArr[i] = suffixes[i].index; } return suffixArr; } // To construct and return LCP static int [] kasai(String txt, int [] suffixArr) { int n = suffixArr.length; int [] lcp = new int [n]; int [] invSuff = new int [n]; for ( int i = 0 ; i < n; i++) { invSuff[suffixArr[i]] = i; } int k = 0 ; for ( int i = 0 ; i < n; i++) { if (invSuff[i] == n - 1 ) { k = 0 ; continue ; } int j = suffixArr[invSuff[i] + 1 ]; while (i + k < n && j + k < n && txt.charAt(i + k) == txt.charAt(j + k)) { k++; } lcp[invSuff[i]] = k; if (k > 0 ) { k--; } } return lcp; } // A utility function to find the longest common // extension from index - L and index - R static int LCE( int [] lcp, int [] invSuff, int n, int L, int R) { if (L == R) { return n - L; } int low = minVal(invSuff[L], invSuff[R]); int high = maxVal(invSuff[L], invSuff[R]); int length = lcp[low]; for ( int i = low + 1 ; i < high; i++) { if (lcp[i] < length) { length = lcp[i]; } } return length; } // A function to answer queries of the longest common // extension static void LCEQueries(String txt, int [] suffixArr, Query[] q) { int n = txt.length(); int [] invSuff = new int [n]; for ( int i = 0 ; i < n; i++) { invSuff[suffixArr[i]] = i; } int [] lcp = kasai(txt, suffixArr); for (Query query : q) { int L = query.L; int R = query.R; System.out.println( "LCE (" + L + ", " + R + ") = " + LCE(lcp, invSuff, n, L, R)); } } public static void main(String[] args) { String txt = "abbababba" ; int n = txt.length(); Query[] queries = { new Query( 1 , 2 ), new Query( 1 , 6 ), new Query( 0 , 5 ) }; int [] suffixArr = buildSuffixArray(txt); LCEQueries(txt, suffixArr, queries); } } |
Python3
# Class to represent a query of form (L, R) class Query: def __init__( self , L, R): self .L = L self .R = R # Class to store information of a suffix class Suffix: def __init__( self , index): self .index = index self .rank = [ 0 , 0 ] # Initial ranks of the suffixes # Function to build the suffix array for a given string def build_suffix_array(txt): n = len (txt) # Create a list of suffixes suffixes = [Suffix(i) for i in range (n)] for i in range (n): # Rank of suffixes using first two characters suffixes[i].rank[ 0 ] = ord (txt[i]) - ord ( 'a' ) suffixes[i].rank[ 1 ] = ord (txt[i + 1 ]) - ord ( 'a' ) if (i + 1 ) < n else - 1 # Sort the suffixes using the comparison function suffixes.sort(key = lambda x: (x.rank[ 0 ], x.rank[ 1 ])) ind = [ 0 ] * n # To store indices of suffixes k = 4 while k < 2 * n: rank = 0 prev_rank = suffixes[ 0 ].rank[ 0 ] suffixes[ 0 ].rank[ 0 ] = rank ind[suffixes[ 0 ].index] = 0 for i in range ( 1 , n): # If first rank and next ranks are same as that of previous # suffix in array, assign the same new rank to this suffix if suffixes[i].rank[ 0 ] = = prev_rank and suffixes[i].rank[ 1 ] = = suffixes[i - 1 ].rank[ 1 ]: suffixes[i].rank[ 0 ] = rank else : prev_rank = suffixes[i].rank[ 0 ] rank + = 1 suffixes[i].rank[ 0 ] = rank ind[suffixes[i].index] = i for i in range (n): nextindex = suffixes[i].index + k / / 2 suffixes[i].rank[ 1 ] = suffixes[ind[nextindex]].rank[ 0 ] if nextindex < n else - 1 # Sort the suffixes according to the first k characters suffixes.sort(key = lambda x: (x.rank[ 0 ], x.rank[ 1 ])) k * = 2 # Store indexes of all sorted suffixes in the suffix array suffixArr = [ 0 ] * n for i in range (n): suffixArr[i] = suffixes[i].index return suffixArr # Function to construct and return LCP (Longest Common Prefix) def kasai(txt, suffixArr): n = len (suffixArr) lcp = [ 0 ] * n # To store lcp of all substring pairs invSuff = [ 0 ] * n # To get next substring from suffix array for i in range (n): invSuff[suffixArr[i]] = i k = 0 for i in range (n): if invSuff[i] = = n - 1 : k = 0 continue # Get the next substring from suffix array j = suffixArr[invSuff[i] + 1 ] # To increase k while txt[i+k] and txt[j+k] are equal while i + k < n and j + k < n and txt[i + k] = = txt[j + k]: k + = 1 lcp[invSuff[i]] = k # lcp for the present substring pair # To decrease the value of k for next suffix if k > 0 : k - = 1 return lcp # Function to find the longest common extension from index - L and index - R def LCE(lcp, invSuff, n, L, R): if L = = R: return n - L low = min (invSuff[L], invSuff[R]) high = max (invSuff[L], invSuff[R]) length = lcp[low] for i in range (low + 1 , high): if lcp[i] < length: length = lcp[i] return length # Function to answer queries of the longest common extension def LCEQueries(txt, suffixArr, q): n = len (txt) invSuff = [ 0 ] * n for i in range (n): invSuff[suffixArr[i]] = i lcp = kasai(txt, suffixArr) for query in q: L = query.L R = query.R print (f "LCE ({L}, {R}) = {LCE(lcp, invSuff, n, L, R)}" ) def main(): txt = "abbababba" queries = [Query( 1 , 2 ), Query( 1 , 6 ), Query( 0 , 5 )] suffixArr = build_suffix_array(txt) LCEQueries(txt, suffixArr, queries) if __name__ = = "__main__" : main() |
C#
// Using System namespace for basic I/O operations using System; using System.Linq; // Definition of the Query class representing a query range public class Query { public int L, R; public Query( int L, int R) { this .L = L; this .R = R; } } // Definition of the Suffix class implementing IComparable public class Suffix : IComparable<Suffix> { public int index; public int [] rank = new int [2]; public Suffix( int index) { this .index = index; } public int CompareTo(Suffix other) { return ( this .rank[0] == other.rank[0]) ? this .rank[1].CompareTo(other.rank[1]) : this .rank[0].CompareTo(other.rank[0]); } } // Class implementing the Longest Common Extension algorithm public class LongestCommonExtension { // Function to return the minimum of two integers static int minVal( int x, int y) => (x < y) ? x : y; // Function to return the maximum of two integers static int maxVal( int x, int y) => (x > y) ? x : y; // Function to build the suffix array of a given string static int [] buildSuffixArray( string txt) { int n = txt.Length; Suffix[] suffixes = new Suffix[n]; // Initialize suffix array with ranks for ( int i = 0; i < n; i++) { suffixes[i] = new Suffix(i); suffixes[i].rank[0] = txt[i] - 'a' ; suffixes[i].rank[1] = ((i + 1) < n) ? (txt[i + 1] - 'a' ) : -1; } // Sort the suffix array using the defined comparison Array.Sort(suffixes); int [] ind = new int [n]; // Iterate for all suffixes and update ranks for ( int k = 4; k < 2 * n; k *= 2) { int rank = 0; int prev_rank = suffixes[0].rank[0]; suffixes[0].rank[0] = rank; ind[suffixes[0].index] = 0; for ( int i = 1; i < n; i++) { if (suffixes[i].rank[0] == prev_rank && suffixes[i].rank[1] == suffixes[i - 1].rank[1]) { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = rank; } else { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = ++rank; } ind[suffixes[i].index] = i; } for ( int i = 0; i < n; i++) { int nextindex = suffixes[i].index + k / 2; suffixes[i].rank[1] = (nextindex < n) ? suffixes[ind[nextindex]].rank[0] : -1; } Array.Sort(suffixes); } int [] suffixArr = new int [n]; for ( int i = 0; i < n; i++) { suffixArr[i] = suffixes[i].index; } return suffixArr; } // Function to compute the longest common prefix (LCP) array using Kasai's algorithm static int [] kasai( string txt, int [] suffixArr) { int n = suffixArr.Length; int [] lcp = new int [n]; int [] invSuff = new int [n]; for ( int i = 0; i < n; i++) { invSuff[suffixArr[i]] = i; } int k = 0; for ( int i = 0; i < n; i++) { if (invSuff[i] == n - 1) { k = 0; continue ; } int j = suffixArr[invSuff[i] + 1]; while (i + k < n && j + k < n && txt[i + k] == txt[j + k]) { k++; } lcp[invSuff[i]] = k; if (k > 0) { k--; } } return lcp; } // Function to compute the Longest Common Extension (LCE) given queries static int LCE( int [] lcp, int [] invSuff, int n, int L, int R) { if (L == R) { return n - L; } int low = minVal(invSuff[L], invSuff[R]); int high = maxVal(invSuff[L], invSuff[R]); int length = lcp[low]; for ( int i = low + 1; i < high; i++) { if (lcp[i] < length) { length = lcp[i]; } } return length; } // Function to compute LCE for a set of queries static void LCEQueries( string txt, int [] suffixArr, Query[] q) { int n = txt.Length; int [] invSuff = new int [n]; for ( int i = 0; i < n; i++) { invSuff[suffixArr[i]] = i; } int [] lcp = kasai(txt, suffixArr); // Iterate through each query and compute LCE foreach (Query query in q) { int L = query.L; int R = query.R; Console.WriteLine( "LCE (" + L + ", " + R + ") = " + LCE(lcp, invSuff, n, L, R)); } } // Driver Code public static void Main( string [] args) { // Sample input string string txt = "abbababba" ; int n = txt.Length; // Sample queries Query[] queries = { new Query(1, 2), new Query(1, 6), new Query(0, 5) }; // Build the suffix array int [] suffixArr = buildSuffixArray(txt); // Compute LCE for the given queries LCEQueries(txt, suffixArr, queries); } } |
Javascript
// Definition of the Query class representing a query range class Query { constructor(L, R) { this .L = L; this .R = R; } } // Definition of the Suffix class implementing Comparable class Suffix { constructor(index) { this .index = index; this .rank = [0, 0]; } compareTo(other) { return ( this .rank[0] === other.rank[0]) ? this .rank[1] - other.rank[1] : this .rank[0] - other.rank[0]; } } // Class implementing the Longest Common Extension algorithm class LongestCommonExtension { // Function to return the minimum of two integers static minVal(x, y) { return (x < y) ? x : y; } // Function to return the maximum of two integers static maxVal(x, y) { return (x > y) ? x : y; } // Function to build the suffix array of a given string static buildSuffixArray(txt) { const n = txt.length; const suffixes = Array.from({ length: n }, (_, i) => new Suffix(i)); // Initialize suffix array with ranks for (let i = 0; i < n; i++) { suffixes[i].rank[0] = txt[i].charCodeAt(0) - 'a' .charCodeAt(0); suffixes[i].rank[1] = (i + 1 < n) ? (txt[i + 1].charCodeAt(0) - 'a' .charCodeAt(0)) : -1; } // Sort the suffix array using the defined comparison suffixes.sort((a, b) => a.compareTo(b)); const ind = new Array(n); // Iterate for all suffixes and update ranks for (let k = 4; k < 2 * n; k *= 2) { let rank = 0; let prev_rank = suffixes[0].rank[0]; suffixes[0].rank[0] = rank; ind[suffixes[0].index] = 0; for (let i = 1; i < n; i++) { if (suffixes[i].rank[0] === prev_rank && suffixes[i].rank[1] === suffixes[i - 1].rank[1]) { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = rank; } else { prev_rank = suffixes[i].rank[0]; suffixes[i].rank[0] = ++rank; } ind[suffixes[i].index] = i; } for (let i = 0; i < n; i++) { const nextindex = suffixes[i].index + k / 2; suffixes[i].rank[1] = (nextindex < n) ? suffixes[ind[nextindex]].rank[0] : -1; } suffixes.sort((a, b) => a.compareTo(b)); } const suffixArr = suffixes.map(suffix => suffix.index); return suffixArr; } // Function to compute the longest common prefix (LCP) array using Kasai's algorithm static kasai(txt, suffixArr) { const n = suffixArr.length; const lcp = new Array(n).fill(0); const invSuff = new Array(n); for (let i = 0; i < n; i++) { invSuff[suffixArr[i]] = i; } let k = 0; for (let i = 0; i < n; i++) { if (invSuff[i] === n - 1) { k = 0; continue ; } const j = suffixArr[invSuff[i] + 1]; while (i + k < n && j + k < n && txt[i + k] === txt[j + k]) { k++; } lcp[invSuff[i]] = k; if (k > 0) { k--; } } return lcp; } // Function to compute the Longest Common Extension (LCE) given queries static LCE(lcp, invSuff, n, L, R) { if (L === R) { return n - L; } const low = LongestCommonExtension.minVal(invSuff[L], invSuff[R]); const high = LongestCommonExtension.maxVal(invSuff[L], invSuff[R]); let length = lcp[low]; for (let i = low + 1; i < high; i++) { if (lcp[i] < length) { length = lcp[i]; } } return length; } // Function to compute LCE for a set of queries static LCEQueries(txt, suffixArr, queries) { const n = txt.length; const invSuff = new Array(n); for (let i = 0; i < n; i++) { invSuff[suffixArr[i]] = i; } const lcp = LongestCommonExtension.kasai(txt, suffixArr); // Iterate through each query and compute LCE for (const query of queries) { const L = query.L; const R = query.R; console.log(`LCE (${L}, ${R}) = ${LongestCommonExtension.LCE(lcp, invSuff, n, L, R)}`); } } // Driver Code static main() { // Sample input string const txt = "abbababba" ; const n = txt.length; // Sample queries const queries = [ new Query(1, 2), new Query(1, 6), new Query(0, 5)]; // Build the suffix array const suffixArr = LongestCommonExtension.buildSuffixArray(txt); // Compute LCE for the given queries LongestCommonExtension.LCEQueries(txt, suffixArr, queries); } } // Run the main method LongestCommonExtension.main(); // This code is contributed by shivamgupta0987654321 |
LCE (1, 2) = 1 LCE (1, 6) = 3 LCE (0, 5) = 4
Analysis of Reduction to RMQ method
Time Complexity :
- To construct the lcp and the suffix array it takes O(N.logN) time.
- To answer each query it takes O(|invSuff[R] ā invSuff[L]|).
Hence the overall time complexity is O(N.logN + Q. (|invSuff[R] ā invSuff[L]|))
where,
Q = Number of LCE Queries.
N = Length of the input string.
invSuff[] = Inverse suffix array of the input string.
Although this may seems like an inefficient algorithm but this algorithm generally outperforms all other algorithms to answer the LCE queries.
We will give a detail description of the performance of this method in the next set.
Auxiliary Space: We use O(N) auxiliary space to store lcp, suffix and inverse suffix arrays.
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