I got to the solution, just was close. Here is what I missed:
"The opponent intends to choose the coin which leaves the user with minimum value."
Here is updated solution:
#include<stdio.h>
int max(int a, int b) {
return a>b?a:b;
}
int min(int a, int b) {
return a<b?a:b;
}
int maxCoins(int arr[], int i, int j, int turn) {
int a,b;
if(i==j) {
if(turn == 1) return arr[i];
else return 0;
}
if(turn) {
a = arr[i] +maxCoins(arr,i+1,j,0);
b = arr[j] + maxCoins(arr,i,j-1,0);
return max(a,b);
} else {
return min(maxCoins(arr,i+1,j,1), maxCoins(arr,i,j-1,1));
}
}
int main() {
int arr[4] = {8, 15, 3, 7 };//{6,7,4,1,10,5,4,9,20,8}; //{2,3,4,5,6,7,8,9,10,11};
printf("%d\n",maxCoins(arr,0,3,1));
}
This link discusses the strategy: http://www.geeksforgeeks.org/dynamic-programming-set-31-optimal-strategy-for-a-game/