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 on 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.
Implementation:
C++
// 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
# 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 code if __name__ == "__main__": string = "banana" K = 10 print(printKthCharInConcatSubstring(string, K))# This code is contributed by sanjeev2552 |
Output
n
This article is contributed by Utkarsh Trivedi. If you like neveropen 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 neveropen main page and help other Geeks
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