George should understand that Lily is not interested in hanging out

I've done it :) Passing all test :) My approach is:

Prep stage:

1. create map with:
   • keys : values from input list,
   • values : indexes of values from input list,
2. make a copy of sorted input list

Algo stage:

Iterate through input list, and compare current value (lets call it cv) against value from sorted list (lets call it scv). If it is diffrent:

• increment number of swaps
• get index of the scv from map - map[scv],
• in input list swap cv with scv,
• update map[cv]=map[scv] (map[scv] does not need to be updated as we are not going to use it any more)

Then you need to execute it on input list and reversed input list - the smaller return value - is the answer.

Time complexity is equal to sort time complexity (usually O(n logn) ). Space O(n)

Finally solved it! Spent some time trying to figure out the hidden details. Would appreciate if there had been another test case. Things i learnt - Java specifics (might be useful if you're stuck):

1. You have to sort the array in both ascending and descending order as both will yield a minimum sum.
2. Remember to use a LinkedHashMap (if you are using one) and not a HashMap to store the element key and index value (to preserve insertion order).
3. A copy of an ArrayList is a shallow copy. You need to addAll(originalArray) to create another array with the same values as the originalArray.

The algorithm is pretty simple: (just a brief run-through , note the details, figure out the implementation)

• compare every element of original array with a copy of the sorted array (refer 3)
• when there's a mismatch, swap elements into their respective positions
• count the total number of swaps
• repeat the above with the array reversed (refer 1)
• use a hashmap as a cache/lookup to speed things up O(1) (refer 2).

Example:

7 3 0 4 1 6 5 2 - original array

0 1 2 3 4 5 6 7 - sorted array

0 3 7 4 1 6 5 2 - after first swap

0 1 7 4 3 6 5 2 - after second swap

0 1 2 4 3 6 5 7 - after third swap

0 1 2 3 4 6 5 7 - after fourth swap

0 1 2 3 4 5 6 7 - after fifth swap

Did you notice the pattern? All the best!

I take issue with the problem. I propose that the solution is to not help Greg in the first place, and ask Lily out myself.

In this problem, my intuition was telling me to just count the different positions between the input array and the sorted input array and then divide this count between 2 rounding up. For example: input array: 1 5 3 4 sorted array: 1 3 4 5 count: 0 1 1 1 = 3 number of swaps = ceil(3/2) = 2

In all the cases that I have tried I get the correct answer. I take into account normal and reverse order. However, test cases #1,#2,#3,#6 give me a wrong answer. Could someone tell me the problem in my reasoning or give me a counter example?