Question

I was practicing the problem on Algorithm games , I tried the following problem but couldn’t find the efficient way to do it::

So can you please help me.Here is the problem.

This is the exact link::
http://community.topcoder.com/tc?module=Static&d1=match_editorials&d2=srm228

Answer 1

I assume that you would like to solve this one by yourself, so I will give you a hint: this problem has optimal substructure.

Imagine that you have both solutions for N-1 coins (without the leftmost one and without the rightmost one). Would it be easy to calculate a solution for N coins then?

You can use two related techniques to exploit this property – dynamic programming and its subtype called memoization. The idea is to store a solution to each sub-problem with L coins missing from the left and R coins missing from the right (use an NxN array for it). Before solving a sub-problem, check the array to see if you’ve already solved it. You would need to solve at most N^2/2 subproblems to arrive at a solution.

Here is pseudocode for a memoized solution:

// solved[L][R] array contains the highest value a player could get
// on a subproblem where L coins are missing from the left
// and R are missing from the right of the original sequence on his move
int solved[N][N] // initialize each element to -1.
int coins[N]     // initialize with coin values 

int solve(int L, int R) {
    if (L+R == N) return 0; // No coins remain
    if (solved[L][R] >= 0) return solved[L][R];
    int remaining = sum(coins from L to R)
    int takeLeft = remaining - solve(L+1, R);
    int takeRight = remaining - solve(L, R+1);
    int result = max(takeLeft, takeRight);
    solved[L][R] = result;
    return result;
}

main() {
    int alice = solve(0, 0);
    int bob = sum(coins) - alice;
}

I remember TopCoder running this problem some time in early 2005 or 2006, but I do not remember enough details to search their problem archive.