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
Above idea is implemented in below code.
// 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 alphabatically 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; } |
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Output:
10
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