# Print Kth character in sorted concatenated substrings of a string

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’
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 dorting. 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/C++ program to print Kth character
// in sorted concatenated substrings
#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;
}

//    Utility method to get sum of first N numbers
int sumOfFirstN(int N)
{
return (N * (N + 1)) / 2;
}

// Returns Kth character in sorted concatenated
// substrings of str
char 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 susbtring */
if (K <= curSubstringLen)
return str[(suffixArr[i] + K - 1)];
else
K -= curSubstringLen;

}
break;
}
else
K -= charToSkip;
}
}

//    Driver code to test above methods
int main()
{
string str = "banana";
int K = 10;
cout << printKthCharInConcatSubstring(str, K);
return 0;
}
```

Output:

```n
```

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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