# Stone Game

Note the problem description in LintCode is different from LeetCode for "Stone Game". Here we discuss the LintCode version:

## Question

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. 1.
At each step of the game, the player can merge two adjacent piles to a new pile.
2. 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 `43`
Tags
Dynamic Programming
Related Problems
Hard Burst Balloons 29 %

## Analysis

### Dynamic Programming

DP四要素
• State:
• `dp[i][j]`表示把第i到第j个石子合并到一起的最小花费
• Function:
• 预处理`sum[i,j]`表示i到j所有石子价值和
• `dp[i][j] = min(dp[i][k]+dp[k+1][j]+sum[i,j])` 对于所有`k`属于`{i,j}`
• Intialize:
• for each i
• `dp[i][i] = 0`
• `dp[0][n-1]`

### Divide & Conquer + Memoization

int cost = memoizedSearch(start, i, A, dp, sum) + memoizedSearch(i + 1, end, A, dp, sum) + sum[end + 1] - sum[start];

`i` ~ `[start, end - 1]`, => `[start, i]`, `[i + 1, end - 1]`
for(int i = start; i < end; i++){
int cost = memoizedSearch(start, i, A, dp, sum) +
memoizedSearch(i + 1, end, A, dp, sum) +
sum[end + 1] - sum[start];
min = Math.min(min, cost);
}
`i` ~ `[start + 1, ... end]`, => `[start, i - 1]`, `[i, end]`
for(int i = start + 1; i <= end; i++){
int cost = memoizedSearch(start, i - 1, A, dp, sum) +
memoizedSearch(i, end, A, dp, sum) +
sum[end + 1] - sum[start];
min = Math.min(min, cost);
}

## Solution

DP - (252 ms)
public class Solution {
/**
* @param A an integer array
* @return an integer
*/
int search(int l, int r, int[][] f, int[][] visit, int[][] sum) {
if(visit[l][r] == 1) {
return f[l][r];
}
if(l == r) {
visit[l][r] = 1;
return f[l][r];
}
f[l][r] = Integer.MAX_VALUE;
for (int k = l; k < r; k++) {
f[l][r] = Math.min(f[l][r], search(l, k, f, visit, sum) + search(k + 1, r, f, visit, sum) + sum[l][r]);
}
visit[l][r] = 1;
return f[l][r];
}
public int stoneGame(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
int n = A.length;
// initialize
int[][] f = new int[n][n];
int[][] visit = new int[n][n];
for (int i = 0; i < n; i++) {
f[i][i] = 0;
}
// preparation
int[][] sum = new int[n][n];
for (int i = 0; i < n; i++) {
sum[i][i] = A[i];
for (int j = i + 1; j < n; j++) {
sum[i][j] = sum[i][j - 1] + A[j];
}
}
return search(0, n-1, f, visit, sum);
}
}
(Preferred) Divide and Conquer + Memoization (251 ms AC)
public class Solution {
/**
* @param A an integer array
* @return an integer
*/
public int stoneGame(int[] A) {
if(A == null || A.length == 0) return 0;
int n = A.length;
// Minimum cost to merge interval dp[i][j]
int[][] dp = new int[n][n];
int[] sum = new int[n + 1];
// Pre-process interval sum
for(int i = 0; i < n; i++){
sum[i + 1] = sum[i] + A[i];
}
return memoizedSearch(0, n - 1, A, dp, sum);
}
private int memoizedSearch(int start, int end, int[] A, int[][] dp, int[] sum){
if(start > end) return 0;
if(start == end) return 0;
if(start + 1 == end) return A[start] + A[end];
if(dp[start][end] != 0) return dp[start][end];
int min = Integer.MAX_VALUE;
for(int i = start; i < end; i++){
int cost = memoizedSearch(start, i, A, dp, sum) +
memoizedSearch(i + 1, end, A, dp, sum) +
sum[end + 1] - sum[start];
min = Math.min(min, cost);
}
dp[start][end] = min;
return min;
}
}