Lexicographical Maximum substring of string
Last Updated :
17 Aug, 2022
Given a string s we have to find the lexicographical maximum substring of a string
Examples:
Input : s = "ababaa"
Output : babaa
Explanation : "babaa" is the maximum lexicographic substring formed from this string
Input : s = "asdfaa"
Output : sdfaa
The idea is simple, we traverse through all substrings. For every substring, we compare it with the current result and update the result if needed.
Below is the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
string LexicographicalMaxString(string str)
{
string mx = "" ;
for ( int i = 0; i < str.length(); ++i)
mx = max(mx, str.substr(i));
return mx;
}
int main()
{
string str = "ababaa" ;
cout << LexicographicalMaxString(str);
return 0;
}
|
Java
class GFG {
static String LexicographicalMaxString(String str)
{
String mx = "" ;
for ( int i = 0 ; i < str.length(); ++i) {
if (mx.compareTo(str.substring(i)) <= 0 ) {
mx = str.substring(i);
}
}
return mx;
}
public static void main(String[] args)
{
String str = "ababaa" ;
System.out.println(LexicographicalMaxString(str));
}
}
|
Python3
def LexicographicalMaxString( str ):
mx = ""
for i in range ( len ( str )):
mx = max (mx, str [i:])
return mx
if __name__ = = '__main__' :
str = "ababaa"
print (LexicographicalMaxString( str ))
|
C#
using System;
public class GFG {
static String LexicographicalMaxString(String str)
{
String mx = "" ;
for ( int i = 0; i < str.Length; ++i) {
if (mx.CompareTo(str.Substring(i)) <= 0) {
mx = str.Substring(i);
}
}
return mx;
}
public static void Main()
{
String str = "ababaa" ;
Console.WriteLine(LexicographicalMaxString(str));
}
}
|
Javascript
<script>
function LexicographicalMaxString(str)
{
var mx = "" ;
for ( var i = 0; i < str.length; ++i) {
if (mx.localeCompare(str.substring(i)) <= 0) {
mx = str.substring(i);
}
}
return mx;
}
var str = "ababaa" ;
document.write(LexicographicalMaxString(str));
</script>
|
Complexity Analysis:
- Time complexity : O(n2)
- Auxiliary Space : O(n)
Optimization: We find the largest character and all its indexes. Now we simply traverse through all instances of the largest character to find lexicographically maximum substring.
Here we follow the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
string LexicographicalMaxString(string str)
{
char maxchar = 'a' ;
vector< int > index;
for ( int i = 0; i < str.length(); i++) {
if (str[i] >= maxchar) {
maxchar = str[i];
index.push_back(i);
}
}
string maxstring = "" ;
for ( int i = 0; i < index.size(); i++) {
if (str.substr(index[i], str.length()) > maxstring) {
maxstring = str.substr(index[i], str.length());
}
}
return maxstring;
}
int main()
{
string str = "acbacbc" ;
cout << LexicographicalMaxString(str);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static String LexicographicalMaxString(String str)
{
char maxchar = 'a' ;
ArrayList<Integer> index = new ArrayList<Integer>();
for ( int i = 0 ; i < str.length(); i++)
{
if (str.charAt(i) >= maxchar)
{
maxchar = str.charAt(i);
index.add(i);
}
}
String maxstring = "" ;
for ( int i = 0 ; i < index.size(); i++)
{
if (str.substring(index.get(i),
str.length()).compareTo( maxstring) > 0 )
{
maxstring = str.substring(index.get(i),
str.length());
}
}
return maxstring;
}
public static void main (String[] args)
{
String str = "acbacbc" ;
System.out.println(LexicographicalMaxString(str));
}
}
|
Python3
def LexicographicalMaxString(st):
maxchar = 'a'
index = []
for i in range ( len (st)):
if (st[i] > = maxchar):
maxchar = st[i]
index.append(i)
maxstring = ""
for i in range ( len (index)):
if (st[index[i]: len (st)] >
maxstring):
maxstring = st[index[i]:
len (st)]
return maxstring
if __name__ = = "__main__" :
st = "acbacbc"
print (LexicographicalMaxString(st))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static string LexicographicalMaxString( string str)
{
char maxchar = 'a' ;
List< int > index = new List< int >();
for ( int i = 0; i < str.Length; i++)
{
if (str[i] >= maxchar)
{
maxchar = str[i];
index.Add(i);
}
}
string maxstring = "" ;
for ( int i = 0; i < index.Count; i++)
{
if (str.Substring(index[i]).CompareTo(maxstring) > 0)
{
maxstring = str.Substring(index[i]);
}
}
return maxstring;
}
static public void Main()
{
string str = "acbacbc" ;
Console.Write(LexicographicalMaxString(str));
}
}
|
Javascript
<script>
function LexicographicalMaxString(str)
{
let maxchar = 'a' ;
let index = [];
for (let i = 0; i < str.length; i++)
{
if (str[i] >= maxchar)
{
maxchar = str[i];
index.push(i);
}
}
let maxstring = "" ;
for (let i = 0; i < index.length; i++)
{
if (str.substring(index[i],
str.length) > maxstring)
{
maxstring = str.substring(index[i],
str.length);
}
}
return maxstring;
}
let str = "acbacbc" ;
document.write(LexicographicalMaxString(str));
</script>
|
Complexity Analysis:
- Time Complexity: O(n*m) where n is the length of the string and m is the size of index array.
- Auxiliary Space: O(n + m) where n is the length of the string and m is the size of index array.
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