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

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!).