248 views
How to generate permutation of a string? What will be the complexity?

retagged | 248 views
0
Can it be done by backtracking?
0
Try recursion. Permute for length l-1, add the last char at each position of these permuted strings.
0
+1
s is the starting index of the string,e is the ending index of the string.

void permute(char *x, int s, int e)
{
int i;
if (s== e)
printf("%s\n", x);
else
{
for (i = s; i <= e; i++)
{
swap((x+s), (x+i));
permute(x, s+1, e);
swap((x+s), (x+i));
}
}
}

There are man methods and best case complexity is $O(n!)$ as there are $n!$ permutations for a string. I'm using Java code as it is recommended for most top tier interviews. (C is not suitable for many interviews as in Google).

Here permute and permute1 are two different permutation methods.

import java.io.*;
import java.util.ArrayList;

public class permuteString{

ArrayList<String> permute(String str){
ArrayList<String> al = new ArrayList<String>();
if(str.length() == 0)
{
return al;
}
ArrayList<String> words = permute(str.substring(1, str.length()));
char    c = str.charAt(0);
for(String word: words)
{
for(int i = 0; i<= word.length(); i++)

}
return al       ;

}

void permute1(String prefix, String s) {
if(s.length() == 0)
{
System.out.println(prefix);
}
for(int i =0; i < s.length(); i++)
{
permute1(prefix+s.charAt(i), s.substring(0,i)+s.substring(i+1, s.length()));
}

}
public static void main(String args[]) {
permuteString mystring = new permuteString(); //Initialize class
mystring.permute1("", args[0]);
System.out.println("____________________________");
ArrayList<String> words =       mystring.permute(args[0]);
for(String word:words)
System.out.println(word);

}

}

by Veteran (424k points)
0
Sir, Is not this can be solved by using Tower of Hanoi recursion.?
0
How?
0
void permute(char *x, int s, int e)
{
int i;
if (s== e)
printf("%s\n", x);
else
{
for (i = s; i <= e; i++)
{
swap((x+s), (x+i));
permute(x, s+1, e);
swap((x+s), (x+i));
}
}
}