# Clash of Kings - Editorial

practice

Difficulty : medium

Pre-requisites : dp over subset, bitmasking.

Solution :

### Explanation by Amit Saharana

I will explain my solution : which is a minimax dp.

Basically, the first king wants to maximize and second king wants to minimize the finals sum of pairs of distances.
Also k <= 12. Maximum number of possible ways in which we can assign some cities to first king = kC(k/2) <= 12C6 <= 1000.

So, the solution is straightforward : dp(cities assigned to first king, cities assigned to second king, number of cities assigned):

• Base case : If number of cities assigned=k -> calculate pairwise distances using information in cities assigned to first king, second king and return that number.

• If number of cities assigned is divisible by 2 => assign it to first king. Choose all remaining cities and take max of recursive calls.

• Else => assign to second king. Take minimum.

Encode assigned cities as bitmasks, and store dp values in a map.

Pre-calculate distances using flyod-warshall.

Time Complexity: O((kC(k/2))^2 * k).

The recursive nature and many overlapping states ensure that the solution passes in time.

You can also have a look at nicely written tester’s code.