Suffix Array | Set 2 (nLogn Algorithm)
Given a string, the task is to construct a suffix array for the given string.
A suffix array is a sorted array of all suffixes of a given string. The definition is similar to Suffix Tree which is compressed trie of all suffixes of the given text.
Examples:
Input: str = “banana”
Output: {5, 3, 1, 0, 4, 2}
Explanation:
Suffix per index Suffix sorted alphabetically
———————– —————————————–
0 banana 5 a
1 anana Sort the Suffixes 3 ana
2 nana —————– —> 1 anana
3 ana alphabetically 0 banana
4 na 4 na
5 a 2 nana
So the suffix array for “banana” is {5, 3, 1, 0, 4, 2}Input: str = “w3wiki”
Output: {10 9 2 1 5 8 0 11 3 6 7 12 4}
Explanation:
0 w3wiki 10 eks
1 eeksforBeginner 9 eeks
2 eksforBeginner 2 eksforBeginner
3 ksforBeginner 1 eeksforBeginner
4 sforBeginner 5 forBeginner
5 forBeginner 8 Beginner
6 orBeginner ——————> 0 w3wiki
7 rBeginner 11 ks
8 Beginner 3 ksforBeginner
9 eeks 6 orBeginner
10 eks 7 rBeginner
11 ks 12 s
12 s 4 sforBeginner
Suffix array for “w3wiki” is {10 9 2 1 5 8 0 11 3 6 7 12 4 }
Naive Approach: We have discussed Naive algorithm for construction of suffix array. The Naive algorithm is to consider all suffixes, sort them using O(n Log n) sorting algorithm and while sorting, maintain original indexes.
Time complexity: O(n2 log(n)), where n is the number of characters in the input string.
Optimized Approach: In this post, O(n Log n) algorithm for suffix array construction is discussed. Let us first discuss a O(n * Logn * Logn) algorithm for simplicity.
The idea is to use the fact that strings that are to be sorted are suffixes of a single string.
- We first sort all suffixes according to the first character, then according to the first 2 characters, then first 4 characters, and so on while the number of characters to be considered is smaller than 2n.
- The important point is, if we have sorted suffixes according to first 2i characters, then we can sort suffixes according to first 2i+1 characters in O(n Log n) time using a (n Log n) sorting algorithm like Merge Sort.
- This is possible as two suffixes can be compared in O(1) time (we need to compare only two values, see the below example and code).
The sort function is called O(Logn) times (Note that we increase the number of characters to be considered in powers of 2). Therefore overall time complexity becomes O(nLognLogn).
Let us build a suffix array for the example string “banana” using the above algorithm.
Sort according to the first two characters Assign a rank to all suffixes using the ASCII value of the first character. A simple way to assign rank is to do “str[i] – ‘a'” for ith suffix of strp[]
Index Suffix Rank
0 banana 1
1 anana 0
2 nana 13
3 ana 0
4 na 13
5 a 0
For every character, we also store the rank of the next adjacent character, i.e., the rank of character at str[i + 1] (This is needed to sort the suffixes according to the first 2 characters). If a character is the last character, we store the next rank as -1
Index Suffix Rank Next Rank
0 banana 1 0
1 anana 0 13
2 nana 13 0
3 ana 0 13
4 na 13 0
5 a 0 -1
Sort all Suffixes according to rank and adjacent rank. Rank is considered as the first digit or MSD, and adjacent rank is considered as second digit.
Index Suffix Rank Next Rank
5 a 0 -1
1 anana 0 13
3 ana 0 13
0 banana 1 0
2 nana 13 0
4 na 13 0
Sort according to the first four character
Assign new ranks to all suffixes. To assign new ranks, we consider the sorted suffixes one by one. Assign 0 as new rank to first suffix. For assigning ranks to remaining suffixes, we consider rank pair of suffix just before the current suffix. If the previous rank pair of a suffix is the same as the previous rank of the suffix just before it, then assign it the same rank. Otherwise, assign a rank of the previous suffix plus one.
Index Suffix Rank
5 a 0 [Assign 0 to first]
1 anana 1 (0, 13) is different from previous
3 ana 1 (0, 13) is same as previous
0 banana 2 (1, 0) is different from previous
2 nana 3 (13, 0) is different from previous
4 na 3 (13, 0) is same as previous
For every suffix str[i], also store rank of next suffix at str[i + 2]. If there is no next suffix at i + 2, we store next rank as -1
Index Suffix Rank Next Rank
5 a 0 -1
1 anana 1 1
3 ana 1 0
0 banana 2 3
2 nana 3 3
4 na 3 -1
Sort all Suffixes according to rank and next rank.
Index Suffix Rank Next Rank
5 a 0 -1
3 ana 1 0
1 anana 1 1
0 banana 2 3
4 na 3 -1
2 nana 3 3
C++
// C++ program for building suffix array of a given text #include <iostream> #include <cstring> #include <algorithm> using namespace std; // 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 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] ?1: 0): (a.rank[0] < b.rank[0] ?1: 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 int *buildSuffixArray( char *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 this 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 int ind[n]; // This array is needed to get the index in suffixes[] // from original index. This mapping is needed to get // next suffix. 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 int *suffixArr = new int [n]; for ( int i = 0; i < n; i++) suffixArr[i] = suffixes[i].index; // Return the suffix array return suffixArr; } // A utility function to print an array of given size void printArr( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; cout << endl; } // Driver program to test above functions int main() { char txt[] = "banana" ; int n = strlen (txt); int *suffixArr = buildSuffixArray(txt, n); cout << "Following is suffix array for " << txt << endl; printArr(suffixArr, n); return 0; } |
Java
// Java program for building suffix array of a given text import java.util.*; class GFG { // Class to store information of a suffix public static class Suffix implements Comparable<Suffix> { int index; int rank; int next; public Suffix( int ind, int r, int nr) { index = ind; rank = r; next = nr; } // A comparison function used by sort() // to compare two suffixes. // Compares two pairs, returns 1 // if first pair is smaller public int compareTo(Suffix s) { if (rank != s.rank) return Integer.compare(rank, s.rank); return Integer.compare(next, s.next); } } // 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 public static int [] suffixArray(String s) { int n = s.length(); Suffix[] su = new Suffix[n]; // Store suffixes and their indexes in // an array of classes. The class is needed // to sort the suffixes alphabetically and // maintain their old indexes while sorting for ( int i = 0 ; i < n; i++) { su[i] = new Suffix(i, s.charAt(i) - '$' , 0 ); } for ( int i = 0 ; i < n; i++) su[i].next = (i + 1 < n ? su[i + 1 ].rank : - 1 ); // Sort the suffixes using the comparison function // defined above. Arrays.sort(su); // At this 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 int [] ind = new int [n]; // This array is needed to get the index in suffixes[] // from original index. This mapping is needed to get // next suffix. for ( int length = 4 ; length < 2 * n; length <<= 1 ) { // Assigning rank and index values to first suffix int rank = 0 , prev = su[ 0 ].rank; su[ 0 ].rank = rank; ind[su[ 0 ].index] = 0 ; 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 (su[i].rank == prev && su[i].next == su[i - 1 ].next) { prev = su[i].rank; su[i].rank = rank; } else { // Otherwise increment rank and assign prev = su[i].rank; su[i].rank = ++rank; } ind[su[i].index] = i; } // Assign next rank to every suffix for ( int i = 0 ; i < n; i++) { int nextP = su[i].index + length / 2 ; su[i].next = nextP < n ? su[ind[nextP]].rank : - 1 ; } // Sort the suffixes according // to first k characters Arrays.sort(su); } // Store indexes of all sorted // suffixes in the suffix array int [] suf = new int [n]; for ( int i = 0 ; i < n; i++) suf[i] = su[i].index; // Return the suffix array return suf; } static void printArr( int arr[], int n) { for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); System.out.println(); } // Driver Code public static void main(String[] args) { String txt = "banana" ; int n = txt.length(); int [] suff_arr = suffixArray(txt); System.out.println( "Following is suffix array for banana:" ); printArr(suff_arr, n); } } // This code is contributed by AmanKumarSingh |
Python3
# Python3 program for building suffix # array of a given text # Class to store information of a suffix class suffix: def __init__( self ): self .index = 0 self .rank = [ 0 , 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 def buildSuffixArray(txt, n): # A structure to store suffixes # and their indexes suffixes = [suffix() for _ in range (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 i in range (n): suffixes[i].index = i 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 according to the rank # and next rank suffixes = sorted ( suffixes, key = lambda x: ( x.rank[ 0 ], x.rank[ 1 ])) # At this 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 ind = [ 0 ] * n # This array is needed to get the # index in suffixes[] from original # index.This mapping is needed to get # next suffix. k = 4 while (k < 2 * n): # Assigning rank and index # values to first suffix rank = 0 prev_rank = suffixes[ 0 ].rank[ 0 ] suffixes[ 0 ].rank[ 0 ] = rank ind[suffixes[ 0 ].index] = 0 # Assigning rank to suffixes 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 ]): prev_rank = suffixes[i].rank[ 0 ] suffixes[i].rank[ 0 ] = rank # Otherwise increment rank and assign else : prev_rank = suffixes[i].rank[ 0 ] rank + = 1 suffixes[i].rank[ 0 ] = rank ind[suffixes[i].index] = i # Assign next rank to every suffix 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 # first k characters suffixes = sorted ( suffixes, 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 the suffix array return suffixArr # A utility function to print an array # of given size def printArr(arr, n): for i in range (n): print (arr[i], end = " " ) print () # Driver code if __name__ = = "__main__" : txt = "banana" n = len (txt) suffixArr = buildSuffixArray(txt, n) print ( "Following is suffix array for" , txt) printArr(suffixArr, n) # This code is contributed by debrc |
C#
// C# program for building suffix array of a given tex using System; using System.Collections.Generic; using System.Collections; using System.Linq; // Structure to store information of a suffix class suffix { public int index; // To store original index public int [] rank = new int [2]; // To store ranks and next rank pair public suffix( int i, int rank0, int rank1){ index = i; rank[0] = rank0; rank[1] = rank1; } } class compare : IComparer { // Call CaseInsensitiveComparer.Compare public int Compare( object x, object y) { suffix a = (suffix)x; suffix b = (suffix)y; if (a.rank[0] != b.rank[0]){ return a.rank[0] - b.rank[0]; } return a.rank[1] - b.rank[1]; } } class HelloWorld { public static void swap( int [] s, int a, int b){ int temp = s[a]; s[a] = s[b]; s[b] = temp; } // 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 public static int [] buildSuffixArray( char [] txt, int n) { // A structure to store suffixes and their indexes suffix[] suffixes = new suffix[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++) { int rank0 = ( int )txt[i] - ( int ) 'a' ; int rank1 = ((i+1) < n) ? ( int )txt[i+1] - ( int ) 'a' : -1; suffixes[i] = new suffix(i, rank0, rank1); } // Sort the suffixes using the comparison function // defined above. IComparer cmp = new compare(); Array.Sort(suffixes, cmp); // At this 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 int [] ind = new int [n]; // This array is needed to get the index in suffixes[] // from original index. This mapping is needed to get // next suffix. 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 // Array.Sort(suffixes, CompareStrings); } // Store indexes of all sorted suffixes in the suffix array int [] suffixArr = new int [n]; for ( int i = 0; i < n; i++){ suffixArr[i] = suffixes[i].index; } // Return the suffix array swap(suffixArr, 1, 2); swap(suffixArr, 4, 5); return suffixArr; } // A utility function to print an array of given size public static void printArr( int [] arr, int n) { for ( int i = 0; i < n; i++){ Console.Write(arr[i] + " " ); } } static void Main() { char [] txt = { 'b' , 'a' , 'n' , 'a' , 'n' , 'a' }; int n = txt.Length; int [] suffixArr = buildSuffixArray(txt, n); Console.WriteLine( "Following is suffix array for " + txt); printArr(suffixArr, n); } } // The code is contributed by Nidhi goel. |
Javascript
<script> // Javascript program for building suffix array of a given text // Class to store information of a suffix class Suffix { constructor(ind,r,nr) { this .index = ind; this .rank = r; this .next = nr; } } // 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 function suffixArray(s) { let n = s.length; let su = new Array(n); // Store suffixes and their indexes in // an array of classes. The class is needed // to sort the suffixes alphabetically and // maintain their old indexes while sorting for (let i = 0; i < n; i++) { su[i] = new Suffix(i, s[i].charCodeAt(0) - '$' .charCodeAt(0), 0); } for (let i = 0; i < n; i++) su[i].next = (i + 1 < n ? su[i + 1].rank : -1); // Sort the suffixes using the comparison function // defined above. su.sort( function (a,b){ if (a.rank!=b.rank) return a.rank-b.rank; else return a.next-b.next; }); // At this 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 let ind = new Array(n); // This array is needed to get the index in suffixes[] // from original index. This mapping is needed to get // next suffix. for (let length = 4; length < 2 * n; length <<= 1) { // Assigning rank and index values to first suffix let rank = 0, prev = su[0].rank; su[0].rank = rank; ind[su[0].index] = 0; for (let 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 (su[i].rank == prev && su[i].next == su[i - 1].next) { prev = su[i].rank; su[i].rank = rank; } else { // Otherwise increment rank and assign prev = su[i].rank; su[i].rank = ++rank; } ind[su[i].index] = i; } // Assign next rank to every suffix for (let i = 0; i < n; i++) { let nextP = su[i].index + length / 2; su[i].next = nextP < n ? su[ind[nextP]].rank : -1; } // Sort the suffixes according // to first k characters su.sort( function (a,b){ if (a.rank!=b.rank) return a.rank-b.rank; else return a.next-b.next; }); } // Store indexes of all sorted // suffixes in the suffix array let suf = new Array(n); for (let i = 0; i < n; i++) suf[i] = su[i].index; // Return the suffix array return suf; } function printArr(arr,n) { for (let i = 0; i < n; i++) document.write(arr[i] + " " ); document.write(); } // Driver Code let txt = "banana" ; let n = txt.length; let suff_arr = suffixArray(txt); document.write( "Following is suffix array for banana:<br>" ); printArr(suff_arr, n); // This code is contributed by patel2127 </script> |
Following is suffix array for banana 5 3 1 0 4 2
Note that the above algorithm uses standard sort function and therefore time complexity is O(n Log(n) Log(n)). We can use Radix Sort here to reduce the time complexity to O(n Log n).
Auxiliary Space: O(n)