## Understanding how to generate permutation in Java

Given a set of distinct numbers, find all of its permutations.

Permutation is defined as the re-arranging of the elements of the set. For example, {1, 2, 3} has the following six permutations:

1. {1, 2, 3}
2. {1, 3, 2}
3. {2, 1, 3}
4. {2, 3, 1}
5. {3, 1, 2}
6. {3, 2, 1}

If a set has ‘n’ distinct elements it will have n! permutations.

Example 1:

Input: [1,3,5]
Output: [1,3,5], [1,5,3], [3,1,5], [3,5,1], [5,1,3], [5,3,1]

Let’s take the example-1 mentioned above to generate all the permutations. Following a BFS approach, we will consider one number at a time:

1. If the given set is empty then we have only an empty permutation set: []
2. Let’s add the first element (1), the permutations will be: 
3. Let’s add the second element (3), the permutations will be: [3,1], [1,3]
4. Let’s add the third element (5), the permutations will be: [5,3,1], [3,5,1], [3,1,5], [5,1,3], [1,5,3], [1,3,5]

Let’s analyze the permutations in the 3rd and 4th step. How can we generate permutations in the 4th step from the permutations of the 3rd step?

If we look closely, we will realize that when we add a new number (5), we take each permutation of the previous step and insert the new number in every position to generate the new permutations. For example, inserting ‘5’ in different positions of [3,1] will give us the following permutations:

1. Inserting ‘5’ before ‘3’: [5,3,1]
2. Inserting ‘5’ between ‘3’ and ‘1’: [3,5,1]
3. Inserting ‘5’ after ‘1’: [3,1,5]

Here is the visual representation of this algorithm: Here’s the non-recursive version:

  public static List<List> findPermutations(int[] nums) {
List<List> result = new ArrayList();
for (int currentNumber : nums) {
// we will take all existing permutations and add the current number to create new permutations
int n = permutations.size();
for (int i = 0; i < n; i++) {
List oldPermutation = permutations.poll();
// create a new permutation by adding the current number at every position
for (int j = 0; j <= oldPermutation.size(); j++) {
List newPermutation = new ArrayList(oldPermutation);
if (newPermutation.size() == nums.length)
else
}
}
}
return result;
}


and its recursive version:

  public static List<List> generatePermutations(int[] nums) {
List<List> result = new ArrayList();
generatePermutationsRecursive(nums, 0, new ArrayList(), result);
return result;
}

private static void generatePermutationsRecursive(int[] nums, int index, List currentPermutation,
List<List> result) {
if (index == nums.length) {