Counting all permutations of a string (Cracking the Coding Interview, Chapter VI - Example 12)

Question

In Gayle Laakman\'s book \"Cracking the Coding Interview\", chapter VI (Big O), example 12, the problem states that given the following Java code for computing a string\'s permu

Answer 1

According to your video where we have string with 3 characters (ABC), the number of permutations is 6 = 3!, and 6 happens to be equal to 1 + 2 + 3. However, if we have a string with 4 characters (ABCD), the number of permutations should be 4 * 3! as D could be in any position from 1 to 4. With each position of D you can generate 3! permutations for the rest. If you re-draw the tree and count the number of permutations you will see the difference.

According to your code, we have n! = str.length()! permutations, but in each call of the permutations, you also run a loop from 0 to n-1. Therefore, you have O(n * n!).

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Update in response to the edited question

Firstly, in programming, we often have either 0->n-1 or 1->n not 0->n.

Secondly, we don't count the number of nodes in this case as if you take a look at the recursion tree in the clip again, you will see nodes duplicated. The permutations in this case should be the number of leaves which are unique among each other.

For instance, if you have a string with 4 characters, the number of leaves should be 4 * 3! = 24 and it would be the number of permutations. However, in your code snippet, you also have a 0->n-1 = 0->3 loop in each permutation, so you need to count the loops in. Thus, your code complexity in this case is O(n *n!) = O(4 * 4!).

Answer 2

You're right about the number of nodes. That formula gives the exact number, but the method in the book counts some multiple times.

Your sum also seems to be approach e * n! for large n, so can be simplified to O(n!).

It's still technically correct to say the number of calls is no more than n * n!, as this is a valid upper bound. Depending on how this is used, this can be fine, and may be easier prove.

For the time complexity, we need to multiply by the average work done for each node.

First, check the String concatenation. Each iteration creates 2 new Strings to pass to the next node. The length of one String increases by 1, and the length of the other decreases by 1, but the total length is always n, giving a time complexity of O(n) for each iteration.

The number of iterations varies for each level, so we can't just multiply by n. Instead look at the total number of iterations for the whole tree, and get the average for each node. With n = 3:

• The 1 node in the first level iterates 3 times: 1 * 3 = 3
• The 3 nodes in the second level iterate 2 times: 3 * 2 = 6
• The 6 nodes in the third level iterate 1 time: 6 * 1 = 6

The total number of iterations is: 3 + 6 + 6 = 15. This is about the same as number of nodes in the tree. So the average number of iterations for each node is constant.

In total, we have O(n!) iterations that each do O(n) work giving a total time complexity of O(n * n!).