dynamica programming coins in line

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one
of the ends of the line until there are no more coins left. The player with the larger amount of money wins.

Would you rather go first or second? Does it matter?

Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.

Let us look one extra step ahead this time by considering the two coins the opponent will possibly take, Ai+1 and Aj.
If the opponent takes Ai+1, the remaining coins are { Ai+2 …
Aj }, which our maximum is denoted by P(i+2, j). On the other hand, if the opponent takes Aj,
our maximum is P(i+1, j-1). Since the opponent is as smart as you, he would have chosen the choice that yields the minimum amount to you.

Therefore, the maximum amount you can get when you choose Ai is:

P1 = Ai + min { P(i+2, j), P(i+1, j-1) }

Similarly, the maximum amount you can get when you choose Aj is:

P2 = Aj + min { P(i+1, j-1), P(i, j-2) }

Therefore,

P(i, j) = max { P1, P2 }
= max { Ai + min { P(i+2, j),   P(i+1, j-1) },
Aj + min { P(i+1, j-1), P(i,   j-2) } }

Although the above recurrence relation could be implemented in few lines of code, its complexity is exponential. The reason is that each recursive call branches into a total of four separate recursive calls, and it could be n levels
deep from the very first call). Memoization provides an efficient way by avoiding re-computations using intermediate results stored in a table. Below is the code which runs in O(n2)
time and takes O(n2)
space.

Edit:

Updated code with a new function printMoves which prints out all the moves you and the opponent make (assuming both of you are taking the coins in an optimal way).

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const int MAX_N = 100; void printMoves(int P[][MAX_N], int A[], int N) { int sum1 = 0, sum2 = 0; int m = 0, n = N-1; bool myTurn = true; while (m <= n) { int P1 = P[m+1][n]; // If take A[m], opponent can get... int P2 = P[m][n-1]; // If take A cout << (myTurn ? "I" : "You") << " take coin no. "; if (P1 <= P2) { cout << m+1 << " (" << A[m] << ")"; m++; } else { cout << n+1 << " (" << A[n] << ")"; n--; } cout << (myTurn ? ", " : ".\n"); myTurn = !myTurn; } cout << "\nThe total amount of money (maximum) I get is " << P[0][N-1] << ".\n"; } int maxMoney(int A[], int N) { int P[MAX_N][MAX_N] = {0}; int a, b, c; for (int i = 0; i < N; i++) { for (int m = 0, n = i; n < N; m++, n++) { assert(m < N); assert(n < N); a = ((m+2 <= N-1) ? P[m+2][n] : 0); b = ((m+1 <= N-1 && n-1 >= 0) ? P[m+1][n-1] : 0); c = ((n-2 >= 0) ? P[m][n-2] : 0); P[m][n] = max(A[m] + min(a,b), A[n] + min(b,c)); } } printMoves(P, A, N); return P[0][N-1]; }