There is a stone game.At the beginning of the game the player picks n piles of stones in a line.
The goal is to merge the stones in one pile observing the following rules:
1. At each step of the game, the player can merge two adjacent piles to a new pile. 2. The score is the number of stones in the new pile.
You are to determine the minimum of the total score.
Example
For [4, 1, 1, 4], in the best solution, the total score is 18:
1. Merge second and third piles => [4, 2, 4], score +2
2. Merge the first two piles => [6, 4]，score +6
3. Merge the last two piles => [10], score +10
Other two examples:
[1, 1, 1, 1] return 8
[4, 4, 5, 9] return 18
Solution
Bottom up solution. Space complexity O(n^2), time complexity O(n^3).
Basic idea: //dp[k,i] represent min merge from stone[k] to stone[i]
// X X X X | X X X... X X X A // k k+1 i
// dp[0,i] = sum(0,i)+min(dp[0,k]+dp[k+1,i]), for k=0,...,i-1
// X X X X | X X - X ... X X X A
// k j j+1 i
// dp[k+1,i] = sum(k+1,i)+ min(dp[k+1,j]+dp[j+1,i]) for j=i,...k+1
public int minCost(int[] stones) {
int len = stones.length;
if(len==0) return 0;
sumArr = new int[len+1];
for(int i=0; i<len; i++) {
sumArr[i+1] = sumArr[i]+stones[i];
}
int[][] dp = new int[len][len];
for(int i=1; i<len; i++) {
int outMin = Integer.MAX_VALUE;//stone[i]
for(int k=i-1; k>=0; k--) {
int min = Integer.MAX_VALUE;
if(k==i-1) min = 0;
for(int j=i-1; j>=k+1; j--) {
min = Math.min(min,sum(k+1,i)+dp[k+1][j]+dp[j+1][i]);
}
dp[k+1][i] = min;
outMin = Math.min(outMin,sum(0,i)+dp[0][k]+dp[k+1][i]);
}
dp[0][i] = outMin;
}
return dp[0][len-1];
}
public int sum(int i, int j) {
return sumArr[j+1]-sumArr[i];
}
