Print Kth character in sorted concatenated substrings of a string - GeeksforGeeks

29 Dec, 2021
Given a string of lower alphabetic characters, find K-th character in a string formed by substrings (of given string) when concatenated in sorted form. 
Examples:  
Input : str = “banana”
          K = 10
Output : n
All substring in sorted form are,
"a", "an", "ana", "anan", "anana", 
"b", "ba", "ban", "bana", "banan", 
"banana", "n", "na", "nan", "nana"
Concatenated string = “aananaanana
nanabbabanbanabananbananannanannana”
We can see a 10th character in the 
above concatenated string is ‘n’ 
which is our final answer.
A simple solution is to generate all substrings of a given string and store them in an array. Once substrings are generated, sort them and concatenate after sorting. Finally print K-th character in the concatenated string.
An efficient solution is based om counting distinct substring of a string using suffix array. Same method is used in solving this problem also. After getting suffix array and lcp array, we loop over all lcp values and for each such value, we calculate characters to skip. We keep subtracting these many characters from our K, when character to skip becomes more than K, we stop and loop over substrings corresponding to current lcp[i], in which we loop from lcp[i] till the maximum length of string and then print the Kth character. 
C++
Python3
// C++ program to print Kth character// in sorted concatenated substrings#include <bits/stdc++.h>using namespace std; // Structure to store information of a suffixstruct 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 smallerint 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 stringvector<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;} //    Utility method to get sum of first N numbersint sumOfFirstN(int N){    return (N * (N + 1)) / 2;} // Returns Kth character in sorted concatenated// substrings of strchar printKthCharInConcatSubstring(string str, int K){    int n = str.length();    //  calculating suffix array and lcp array    vector<int> suffixArr = buildSuffixArray(str, n);    vector<int> lcp = kasai(str, suffixArr);     for (int i = 0; i < lcp.size(); i++)     {         //    skipping characters common to substring         //    (n - suffixArr[i]) is length of current         //  maximum substring lcp[i] will length of         // common substring        int charToSkip = sumOfFirstN(n - suffixArr[i]) -                         sumOfFirstN(lcp[i]);         /*    if characters are more than K, that means            Kth character belongs to substring            corresponding to current lcp[i]*/        if (K <= charToSkip)        {            // loop from current lcp value to current            // string length            for (int j = lcp[i] + 1; j <= (n-suffixArr[i]); j++)            {                int curSubstringLen = j;                 /* Again reduce K by current substring's                   length one by one and when it becomes less,                    print Kth character of current substring */                if (K <= curSubstringLen)                    return str[(suffixArr[i] + K - 1)];                else                    K -= curSubstringLen;             }            break;        }        else            K -= charToSkip;     }} //    Driver code to test above methodsint main(){    string str = "banana";    int K = 10;    cout << printKthCharInConcatSubstring(str, K);    return 0;}
# Python3 program to print Kth character# in sorted concatenated substrings # Structure to store information of a suffixclass suffix:         def __init__(self):                 self.index = 0                 # To store original index        self.rank = [0] * 2                 # To store ranks and next        # rank pair # 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 stringdef buildSuffixArray(txt: str, n: int) -> list:     # A structure to store suffixes    # and their indexes    suffixes = [0] * n    for i in range(n):        suffixes[i] = suffix()     # 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 using the comparison function    # defined above.    suffixes.sort(key = lambda a: a.rank)     # 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    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:        k *= 2                 # for k in range(4, 2 * n, k * 2):         # 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.sort(key = lambda a : a.rank)     # Store indexes of all sorted suffixes    # in the suffix array    suffixArr = []    for i in range(n):        suffixArr.append(suffixes[i].index)     # Return the suffix array    return suffixArr # To construct and return LCP */def kasai(txt: str, suffixArr: list) -> list:     n = len(suffixArr)     # To store LCP array    lcp = [0] * n     # 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.    invSuff = [0] * n     # Fill values in invSuff[]    for i in range(n):        invSuff[suffixArr[i]] = i     # Initialize length of previous LCP    k = 0     # Process all suffixes one by one    # starting from first suffix in txt[]    for i in range(n):         # 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        j = suffixArr[invSuff[i] + 1]         # Directly start matching from k'th index as        # at-least k-1 characters will match        while (i + k < n and j + k < n and           txt[i + k] == txt[j + k]):            k += 1         lcp[invSuff[i]] = k        # lcp for the present suffix.         # Deleting the starting character        # from the string.        if (k > 0):            k -= 1     # Return the constructed lcp array    return lcp # Utility method to get sum of first N numbersdef sumOfFirstN(N: int) -> int:     return (N * (N + 1)) // 2 # Returns Kth character in sorted concatenated# substrings of strdef printKthCharInConcatSubstring(string: str,                               K: int) -> str:     n = len(string)         # Calculating suffix array and lcp array    suffixArr = buildSuffixArray(string, n)    lcp = kasai(string, suffixArr)     for i in range(len(lcp)):         # Skipping characters common to substring        # (n - suffixArr[i]) is length of current        # maximum substring lcp[i] will length of        # common substring        charToSkip = (sumOfFirstN(n - suffixArr[i]) -                                sumOfFirstN(lcp[i]))         # If characters are more than K, that means        # Kth character belongs to substring        # corresponding to current lcp[i]        if (K <= charToSkip):             # Loop from current lcp value to current            # string length            for j in range(lcp[i] + 1,               (n - suffixArr[i]) + 1):                curSubstringLen = j                 # Again reduce K by current substring's                # length one by one and when it becomes less,                # print Kth character of current substring                if (K <= curSubstringLen):                    return string[(suffixArr[i] + K - 1)]                else:                    K -= curSubstringLen                                 break         else:            K -= charToSkip # Driver codeif __name__ == "__main__":     string = "banana"    K = 10         print(printKthCharInConcatSubstring(string, K)) # This code is contributed by sanjeev2552
Output: 
n
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