LintCode "Coins in a Line III" !!
https://codesolutiony.wordpress.com/2015/05/24/lintcode-coins-in-a-line-iii/
A very juicy one! Deserve more consideration.
class Solution {
public:
/**
* @param values: a vector of integers
* @return: a boolean which equals to true if the first player will win
*/
bool firstWillWin(vector<int> &values) {
int n = values.size();
if (n < ) return true; // Step 1: Find Variables
// since it is 2-end a single dim i is not enough, so - i, j (left, right)
// and dp[i][j] is the max value Play1 can get in [i, j]
vector<vector<int>> dp(n, vector<int>(n)); // Step 2: Choice strategy
// Apparently in every step, we have 2 choices: pick left or right
// so dp[i][j] should depends on dp[i-1][j] and dp[i][j-1]
//
// Step 2.5 Game Thoery
// In each choice at step2, we always Play2 always made the optimal
// choice in last step
//
// Equation
// dp[i][j] = max(
// values[i] + sum[i+1][j] - dp[i+1][j],
// values[j] + sum[i][j-1] - dp[i][j-1]
// ); vector<int> presum(n);
partial_sum(values.begin(), values.end(), presum.begin()); for(int j = ; j < n; j ++)
for(int i = j; i >= ; i --)
{
if (i == j)
{
dp[i][j] = values[i];
}
else
{
int sumij = presum[j] - (i > ? presum[i - ] : );
dp[i][j] = sumij - min(dp[i+][j], dp[i][j-]);
}
} return dp[][n-] > (presum.back() - dp[][n-]);
}
};
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