Hackerrank -- Ashton and string (suffix array)

Question Link

Ashton appeared for a job interview and is asked the following question. arrange all the distinct substrings of a given string in lexicographical order and concatenate them. print the kth character of the concatenated string. it is assured that given value of K will be valid I. e. there will be a kth character. can you help Ashton out with this?

Input Format 
First line will contain a number t I. e. Number of test cases.
First line of each test case will contain a string containing characters (A −z) and second line will contain a number K.

Output Format 
Print kth character (the string is 1 indexed)

Constraints 
1 ≤ T ≤ 5
1 ≤ length ≤ 105
K will be an appropriate integer.

Sample input #00

1dbac3

Sample output #00

c

Explanation #00

The substrings when arranged in Lexicographic Order are as follows

a, ac, b, ba, bac, c, d, db, dba, dbac

On concatenating them, we get

aacbbabaccddbdbadbac

The third character in this string iscAnd hence the answer.

 

Question: Give a string, concatenate all the strings in the lexicographically ascending order to get a large string. Evaluate the K character of the string.

Train of Thought: the result after the strings of this string are sorted alphabetically is actually strings numbered by LCP in the suffix array...

Accepted code:

 1 #include <string> 2 #include <iostream> 3 #include <algorithm> 4 using namespace std; 5  6 const int MAX_N = 100005; 7 typedef long long LL; 8 int n, k, sa[MAX_N], rk[MAX_N], lcp[MAX_N], tmp[MAX_N]; 9 10 bool compare_sa(int i, int j) {11     if (rk[i] != rk[j]) return rk[i] < rk[j];12     int ri = i + k <= n ? rk[i + k] : -1;13     int rj = j + k <= n ? rk[j + k] : -1;14     return ri < rj;15 }16 17 void construct_sa(const string &S, int *sa) {18     n = S.length();19     for (int i = 0; i <= n; i++) {20         sa[i] = i;21         rk[i] = (i < n ? S[i] : -1);22     }23     24     for (k = 1; k <= n; k *= 2) {25         sort(sa, sa + n + 1, compare_sa);26         27         tmp[sa[0]] = 0;28         for (int i = 1; i <= n; i++) {29             tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);30         }31         for (int i = 0; i <= n; i++) rk[i] = tmp[i];32     }33 }34 35 void construct_lcp(const string &S, int *sa, int *lcp) {36     n = S.length();37     for (int i = 0; i <= n; i++) rk[sa[i]] = i;38     39     int h = 0;40     lcp[0] = 0;41     for (int i = 0; i < n; i++) {42         int j = sa[rk[i] - 1];43         44         if (h > 0) h--;45         for (; i + h < n && j + h < n; h++) if (S[i + h] != S[j + h]) break;46         47         lcp[rk[i] - 1] = h;48     }49 } 50 51 string S;52 53 void solve(LL k) {54     n = S.length();55     construct_sa(S, sa);56     construct_lcp(S, sa, lcp);   57     58     for (int i = 0; i < n; i++) {59         int L = lcp[i];60         int left = n - sa[i + 1];61         LL sum = (L + 1 + left) * (LL)(left - L) / 2;62         if (k > sum) {k -= sum;}63         else {64             for (int j = L + 1; j <= left; j++) {65                 if (k <= j) {66                    int index = sa[i + 1] + k;67                    cout << S[index - 1] << endl;68                    return ;69                 } else {70                     k -= j;71                 }72             }73         }74     }75 }76 int main(void) {77     ios::sync_with_stdio(false);78     int T;79     cin >> T;80     while (T--) {81         LL k;82         cin >> S >> k;83         solve(k);84     }85     return 0;86 }

 

Hackerrank -- Ashton and string (suffix array)