Given string str of length N, the task is to obtain the lexicographically largest string by at most one swap.
Note: The swapping characters might not be adjacent.
Examples:
Input: str = “string”
Output: tsring
Explanation:
Lexicographically largest string obtained by swapping string -> tsring.Input: str = “zyxw”
Output: zyxw
Explanation:
The given string is already lexicographically largest
Approach:
To solve the above-mentioned problem, the main idea is to use Sorting and compute the largest lexicographical string possible for the given string. After sorting the given string in descending order, find the first unmatched character from the given string and replace it with the last occurrence of the unmatched character in the sorted string.
Illustration:
str = “geeks”
Sorted string in descending order = “skgee”.
The first unmatched character is in the first place. This character needs to be swapped with the character at this position in the sorted string which results in the lexicographically largest string. On replacing “g” with the “s”, the string obtained is “seekg” which is lexicographically largest after one swap.
Below is the implementation of the above approach:
// C++ implementation to find the // lexicographically largest string // by atmost at most one swap #include <bits/stdc++.h> using namespace std;
// Function to return the // lexicographically largest // string possible by swapping // at most one character string findLargest(string s) { int len = s.size();
// Stores last occurrence
// of every character
int loccur[26];
// Initialize with -1 for
// every character
memset (loccur, -1, sizeof (loccur));
for ( int i = len - 1; i >= 0; --i) {
// Keep updating the last
// occurrence of each character
int chI = s[i] - 'a' ;
// If a previously unvisited
// character occurs
if (loccur[chI] == -1) {
loccur[chI] = i;
}
}
// Stores the sorted string
string sorted_s = s;
sort(sorted_s.begin(), sorted_s.end(),
greater< int >());
for ( int i = 0; i < len; ++i) {
if (s[i] != sorted_s[i]) {
// Character to replace
int chI = sorted_s[i] - 'a' ;
// Find the last occurrence
// of this character
int last_occ = loccur[chI];
// Swap this with the last
// occurrence
swap(s[i], s[last_occ]);
break ;
}
}
return s;
} // Driver Program int main()
{ string s = "yrstvw" ;
cout << findLargest(s);
return 0;
} |
// Java implementation to find the // lexicographically largest string // by atmost at most one swap import java.util.*;
import java.lang.*;
class GFG{
// Function to return the // lexicographically largest // string possible by swapping // at most one character static String findLargest(StringBuilder s)
{ int len = s.length();
// Stores last occurrence
// of every character
int [] loccur = new int [ 26 ];
// Initialize with -1 for
// every character
Arrays.fill(loccur, - 1 );
for ( int i = len - 1 ; i >= 0 ; --i)
{
// Keep updating the last
// occurrence of each character
int chI = s.charAt(i) - 'a' ;
// If a previously unvisited
// character occurs
if (loccur[chI] == - 1 )
{
loccur[chI] = i;
}
}
// Stores the sorted string
char [] sorted_s = s.toString().toCharArray();
Arrays.sort(sorted_s);
reverse(sorted_s);
for ( int i = 0 ; i < len; ++i)
{
if (s.charAt(i) != sorted_s[i])
{
// Character to replace
int chI = sorted_s[i] - 'a' ;
// Find the last occurrence
// of this character
int last_occ = loccur[chI];
// Swap this with the last
// occurrence
char tmp = s.charAt(i);
s.setCharAt(i, s.charAt(last_occ));
s.setCharAt(last_occ, tmp);
break ;
}
}
return s.toString();
} // Function to reverse array static void reverse( char a[])
{ int i, n = a.length;
for (i = 0 ; i < n / 2 ; i++)
{
char t = a[i];
a[i] = a[n - i - 1 ];
a[n - i - 1 ] = t;
}
} // Driver Code public static void main(String[] args)
{ StringBuilder s = new StringBuilder( "yrstvw" );
System.out.println(findLargest(s));
} } // This code is contributed by offbeat |
# Python3 implementation to find the # lexicographically largest string # by atmost at most one swap # Function to return the # lexicographically largest # string possible by swapping # at most one character def findLargest(s):
Len = len (s)
# Stores last occurrence
# of every character
# Initialize with -1 for
# every character
loccur = [ - 1 for i in range ( 26 )]
for i in range ( Len - 1 , - 1 , - 1 ):
# Keep updating the last
# occurrence of each character
chI = ord (s[i]) - ord ( 'a' )
# If a previously unvisited
# character occurs
if (loccur[chI] = = - 1 ):
loccur[chI] = i
# Stores the sorted string
sorted_s = sorted (s, reverse = True )
for i in range ( Len ):
if (s[i] ! = sorted_s[i]):
# Character to replace
chI = ( ord (sorted_s[i]) -
ord ( 'a' ))
# Find the last occurrence
# of this character
last_occ = loccur[chI]
temp = list (s)
# Swap this with the last
# occurrence
temp[i], temp[last_occ] = (temp[last_occ],
temp[i])
s = "".join(temp)
break
return s
# Driver code s = "yrstvw"
print (findLargest(s))
# This code is contributed by avanitrachhadiya2155 |
// C# implementation to find the // lexicographically largest string // by atmost at most one swap using System;
using System.Collections.Generic;
class GFG{
// Function to return the // lexicographically largest // string possible by swapping // at most one character static string findLargest( char [] s)
{ int len = s.Length;
// Stores last occurrence
// of every character
int [] loccur = new int [26];
// Initialize with -1 for
// every character
Array.Fill(loccur, -1);
for ( int i = len - 1; i >= 0; --i)
{
// Keep updating the last
// occurrence of each character
int chI = s[i] - 'a' ;
// If a previously unvisited
// character occurs
if (loccur[chI] == -1)
{
loccur[chI] = i;
}
}
// Stores the sorted string
char [] sorted_s = ( new string (s)).ToCharArray();
Array.Sort(sorted_s);
Array.Reverse(sorted_s);
for ( int i = 0; i < len; ++i)
{
if (s[i] != sorted_s[i])
{
// Character to replace
int chI = sorted_s[i] - 'a' ;
// Find the last occurrence
// of this character
int last_occ = loccur[chI];
// Swap this with the last
// occurrence
char temp = s[i];
s[i] = s[last_occ];
s[last_occ] = temp;
break ;
}
}
return ( new string (s));
} // Driver Code static void Main()
{ string str = "yrstvw" ;
char [] s = str.ToCharArray();
Console.WriteLine(findLargest(s));
} } // This code is contributed by divyesh072019 |
<script> // Javascript implementation to find the
// lexicographically largest string
// by atmost at most one swap
// Function to return the
// lexicographically largest
// string possible by swapping
// at most one character
function findLargest(s)
{
let len = s.length;
// Stores last occurrence
// of every character
let loccur = new Array(26);
// Initialize with -1 for
// every character
loccur.fill(-1);
for (let i = len - 1; i >= 0; --i)
{
// Keep updating the last
// occurrence of each character
let chI = s[i].charCodeAt() - 'a' .charCodeAt();
// If a previously unvisited
// character occurs
if (loccur[chI] == -1)
{
loccur[chI] = i;
}
}
// Stores the sorted string
let sorted_s = s.join( "" ).split( '' );
sorted_s.sort();
sorted_s.reverse();
for (let i = 0; i < len; ++i)
{
if (s[i] != sorted_s[i])
{
// Character to replace
let chI = sorted_s[i].charCodeAt() - 'a' .charCodeAt();
// Find the last occurrence
// of this character
let last_occ = loccur[chI];
// Swap this with the last
// occurrence
let temp = s[i];
s[i] = s[last_occ];
s[last_occ] = temp;
break ;
}
}
return (s.join( "" ));
}
let str = "yrstvw" ;
let s = str.split( '' );
document.write(findLargest(s));
</script> |
ywstvr
Time Complexity: O(n * log(n)), where n is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.