K-th lexicographically smallest unique substring of a given string
Last Updated :
27 Apr, 2023
Given a string S. The task is to print the K-th lexicographically the smallest one among the different substrings of s.
A substring of s is a string obtained by taking out a non-empty contiguous part in s.
For example, if s = ababc, a, bab and ababc are substrings of s, while ac, z, and an empty string are not. Also, we say that substrings are different when they are different as strings.
Examples:
Input: str = "aba", k = 4
Output: b
All unique substrings are a, ab, aba, b, ba.
Thus the 4th lexicographically smallest substring is b.
Input: str = "geeksforgeeks", k = 5
Output: eeksf
Approach: For an arbitrary string t, each of its proper suffixes is lexicographically smaller than t, and the lexicographic rank of t is at least |t|. Thus, the length of the answer is at most K. Generate all substrings of s whose lengths are at most K. Sort them, unique them, and print the K-th one, where N = |S|.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
void kThLexString(string st, int k, int n)
{
// Set to store the unique substring
set<string> z;
for(int i = 0; i < n; i++)
{
// String to create each substring
string pp;
for(int j = i; j < i + k; j++)
{
if (j >= n)
break;
pp += st[j];
// Adding to set
z.insert(pp);
}
}
// Converting set into a list
vector<string> fin(z.begin(), z.end());
// Sorting the strings int the list
// into lexicographical order
sort(fin.begin(), fin.end());
// Printing kth substring
cout << fin[k - 1];
}
// Driver code
int main()
{
string s = "geeksforgeeks";
int k = 5;
int n = s.length();
kThLexString(s, k, n);
}
// This code is contributed by yatinagg
// Java implementation of
// the above approach
import java.util.*;
class GFG{
public static void kThLexString(String st,
int k, int n)
{
// Set to store the unique substring
Set<String> z = new HashSet<String>();
for(int i = 0; i < n; i++)
{
// String to create each substring
String pp = "";
for(int j = i; j < i + k; j++)
{
if (j >= n)
break;
pp += st.charAt(j);
// Adding to set
z.add(pp);
}
}
// Converting set into a list
Vector<String> fin = new Vector<String>();
fin.addAll(z);
// Sorting the strings int the list
// into lexicographical order
Collections.sort(fin);
// Printing kth substring
System.out.print(fin.get(k - 1));
}
// Driver Code
public static void main(String[] args)
{
String s = "geeksforgeeks";
int k = 5;
int n = s.length();
kThLexString(s, k, n);
}
}
// This code is contributed by divyeshrabadiya07
# Python3 implementation of the above approach
def kThLexString(st, k, n):
# Set to store the unique substring
z = set()
for i in range(n):
# String to create each substring
pp = ""
for j in range(i, i + k):
if (j >= n):
break
pp += s[j]
# adding to set
z.add(pp)
# converting set into a list
fin = list(z)
# sorting the strings int the list
# into lexicographical order
fin.sort()
# printing kth substring
print(fin[k - 1])
s = "geeksforgeeks"
k = 5
n = len(s)
kThLexString(s, k, n)
// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
using System.Collections;
class GFG{
public static void kThLexString(string st,
int k, int n)
{
// Set to store the unique substring
HashSet<string> z = new HashSet<string>();
for(int i = 0; i < n; i++)
{
// String to create each substring
string pp = "";
for(int j = i; j < i + k; j++)
{
if (j >= n)
break;
pp += st[j];
// Adding to set
z.Add(pp);
}
}
// Converting set into a list
ArrayList fin = new ArrayList();
foreach(string s in z)
{
fin.Add(s);
}
// Sorting the strings int the list
// into lexicographical order
fin.Sort();
// Printing kth substring
Console.Write(fin[k - 1]);
}
// Driver Code
public static void Main(string[] args)
{
string s = "geeksforgeeks";
int k = 5;
int n = s.Length;
kThLexString(s, k, n);
}
}
// This code is contributed by rutvik_56
<script>
// JavaScript implementation of the above approach
function kThLexString(st, k, n)
{
// Set to store the unique substring
var z = new Set();
for(var i = 0; i < n; i++)
{
// String to create each substring
var pp = "";
for(var j = i; j < i + k; j++)
{
if (j >= n)
break;
pp += st[j];
// Adding to set
z.add(pp);
}
}
var fin = [];
z.forEach(element => {
fin.push(element);
});
fin.sort();
// Printing kth substring
document.write( fin[k-1]);
}
// Driver code
var s = "geeksforgeeks";
var k = 5;
var n = s.length;
kThLexString(s, k, n);
</script>
Time Complexity: O(nk log(nk) where n is the length of the string, and k is the length of the substring.
Space Complexity: O(nk)
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