Sunday, 23 June 2013

# Algorithm: All permutations of a set

## Problem

Implement a function that gets all possible permutations (or orderings) of the characters in a string. For example for the input string "abc", the output will be "abc", "acb", "bac", "bca", "cab" and "cba".

## Analysis

This problem is very similar to all combinations of a set, though the actual computing of the values will be quite different. Let's start by defining the inputs and outputs.

ArrayList<String> getPermutations(String characters);


Now let's look at how this problem is naturally solved. When I write down a set of permutations by hand, I tend to start with the first letter (a), and then find all permutations without that letter in it. So for "abc" I would write:

a bc
a cb
b ac
b ca
c ab
c ba


This approach can be translated exactly into a recursive function in which for all letters in a string, we pull the letter out of the string and prepend it to all permutations of the string without that letter in it. The base case when the string is a single character will return the character.

permutations(abc) = a + permutations(bc) +
b + permutations(ac) +
c + permutations(ab)

permutations(ab)  = a + permutations(b) +
b + permutations(a)

permutations(a)   = a


## Pseudocode

function getPermutations (string text)
define results as string[]
if text is a single character
return results
foreach char c in text
define innerPermutations as string[]
set innerPermutations to getPermutations (text without c)
foreach string s in innerPermutations
add c + s to results
return results


## Complexity

Much like all combinations of a set, the time and space complexity of the above algorithm should be the same as the number of items produced. The number of unique permutations of any set of size n is n!, therefore our algorithm is O(n!).

## Code

public static ArrayList<String> getPermutations(String text) {
ArrayList<String> results = new ArrayList<String>();

// the base case
if (text.length() == 1) {
return results;
}

for (int i = 0; i < text.length(); i++) {
char first = text.charAt(i);
String remains = text.substring(0, i) + text.substring(i + 1);

ArrayList<String> innerPermutations = getPermutations(remains);

for (int j = 0; j < innerPermutations.size(); j++)