Count of distinct substrings of a string using Suffix Array

Given a string of length n of lowercase alphabet characters, we need to count total number of distinct substrings of this string. 

Examples: 

Input  : str = “ababa”
Output : 10
Total number of distinct substring are 10, which are,
"", "a", "b", "ab", "ba", "aba", "bab", "abab", "baba"
and "ababa"

We have discussed a Suffix Trie based solution in below post : 
Count of distinct substrings of a string using Suffix Trie

We can solve this problem using suffix array and longest common prefix concept. A suffix array is a sorted array of all suffixes of a given string. 
For string “ababa” suffixes are : “ababa”, “baba”, “aba”, “ba”, “a”. After taking these suffixes in sorted form we get our suffix array as [4, 2, 0, 3, 1] 
Then we calculate lcp array using kasai’s algorithm. For string “ababa”, lcp array is [1, 3, 0, 2, 0]
After constructing both arrays, we calculate total number of distinct substring by keeping this fact in mind : If we look through the prefixes of each suffix of a string, we cover all substrings of that string. 

We will explain the procedure for above example,

String  = “ababa”
Suffixes in sorted order : “a”, “aba”, “ababa”,
                            “ba”, “baba”
Initializing distinct substring count by length
of first suffix, 
Count = length(“a”) = 1        
Substrings taken in consideration : “a”

Now we consider each consecutive pair of suffix, 
lcp("a", "aba") = "a".
All characters that are not part of the longest 
common prefix contribute to a distinct substring. 
In the above case, they are 'b' and ‘a'. So they 
should be added to Count.
Count += length(“aba”) - lcp(“a”, “aba”) 
Count  = 3    
Substrings taken in consideration : “aba”, “ab”

Similarly for next pair also,
Count += length(“ababa”) - lcp(“aba”, “ababa”)
Count = 5
Substrings taken in consideration : “ababa”, “abab”

Count += length(“ba”) - lcp(“ababa”, “ba”)
Count = 7
Substrings taken in consideration : “ba”, “b”

Count += length(“baba”) - lcp(“ba”, “baba”)
Count = 9
Substrings taken in consideration : “baba”, “bab”

We finally add 1 for empty string.
count = 10

Implementation:

CPP

// C++ code to count total distinct substrings
// of a string
#include <bits/stdc++.h>
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
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
    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
    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)
{
    int n = suffixArr.size();

    // To store LCP array
    vector<int> lcp(n, 0);

    // 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);

    // 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 don’t
           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;
}

//  method to return count of total distinct substring
int countDistinctSubstring(string txt)
{
    int n = txt.length();
    //  calculating suffix array and lcp array
    vector<int> suffixArr = buildSuffixArray(txt, n);
    vector<int> lcp = kasai(txt, suffixArr);

    // n - suffixArr[i] will be the length of suffix
    // at ith position in suffix array initializing
    // count with length of first suffix of sorted
    // suffixes
    int result = n - suffixArr[0];

    for (int i = 1; i < lcp.size(); i++)

        //  subtract lcp from the length of suffix
        result += (n - suffixArr[i]) - lcp[i - 1];

    result++;  // For empty string
    return result;
}

//  Driver code to test above methods
int main()
{
    string txt = "ababa";
    cout << countDistinctSubstring(txt);
    return 0;
}

Output

10

Time Complexity : O(nlogn), where n is the length of string.
Auxiliary Space : O(n), where n is the length of string.

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.