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# Count of distinct substrings of a string using Suffix Array

• Difficulty Level : Expert
• Last Updated : 06 Jul, 2022

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
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 ``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; i0)``            ``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.

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