A Dynamic programming problem

Question

Can anyone help me find an optimal Dynamic programming algorithm for this problem

On the way to dinner, the CCC competitors are lining up for their delicious curly f

Answer 1

The subproblem is the minimal groups needed to get everyone to dine for a given state of the line. There are a whole lot of possible line states but only a few that will actually be seen, so your memoization should probably be done with a hash map.

Here's some pseudocode.

int dine(string line){
    if(hashmap.contains(line)){
        return hashmap.get(line);
    }
    if(line.length == 0){
        return 0;
    }
    best = N+1;
    for(i=0;i<line.length;i=j){
        type = string[i];
        j = i+1;
        while(type == string[j]){
            j++;
        }
        if(j-i >= K){
             result = dine(string.substring(0,i-1) + string.substring(j,string.length));
             if(result > 0 && result < best){
                  best = result;
             }
        }
    }
    if(best == N+1){
         hashmap.insert(line, -1);
         return -1;
    }
    else {
         hashmap.insert(line, best+1);
         return best+1;
    }
}

If you already found answer for this line, return that answer. If theres no people in the line, you are already done and you don't need to form any more groups.

Assume you can't form any groups. Then try to prove this wrong by trying out all the contiguous groups of like-minded programmers in the line. If the group is big enough to get picked, see how many more moves it would take to get everyone in the resturaunt after removing this group. Keep track of the least moves needed.

If you couldn't find a way to remove all the groups, return -1. Otherwise, return the least moves needed after removing a group plus one to account for the move you made on this step.

Answer 2

I'd prefer not to solve an SPOJ problem in a practical manner for you, so take the following as an existence proof of a poly-time DP.

For K fixed, the set of strings that can dine is context-free. I'm going to use g and h instead of G and H. For example, for K = 3, one grammar looks like

S -> ε | g S g S g S G | h S h S h S H

G -> ε | g S G

H -> ε | h S H

The idea is that either there are no diners, or the first diner dines with at least K - 1 others, between any two of which (and the last and the end) there is a string that can dine.

Now use the weighted variant of CYK to find the minimum-weight parse, where nonempty S productions have weight 1, and all others have weight 0. For K fixed, the running time of CYK is O(N³).

Answer 3

How about divide and conquer? Take a (removable) group somewhere near the middle, and the two groups either side of it, say ...HHHGGGGGHHHHH.... - now there's two possibilities. Either those 2 sets of H's dine in the same group, or they don't. If they dine in the same group, then those G's between them must be removed as a group of precisely those G's (so you may as well try that as your first move). If they don't, then you can solve for the left and right sublists separately. Either case, you've got a shorter list to recurse on.