Stone Game with Recursion

Many days ago I showed how to use dynamic programming to solve the stones game question. see here

Dynamic programming is no doubt a very efficient way to solve this question. However, a recursive algorithm would be more intuitive and easy to understand, hence give better maintainability. With a little hack, e.g. caching, we can still reach a pretty good performance using recursion.

Problem reviewed

from lintcode
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:

At each step of the game, the player can merge two adjacent piles to a new pile.
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

/**
 * GIVEN:
 * A stone game:
 * - a list of stone numbers in a line
 * - you can merge two adjacent stone into a pile
 * - and you can merge two adjacent piles/stones into one bigger pile
 * - each time you merge one pile, you get a score
 *   - score = total score of left pile + total score of right pile
 *
 * ASK:
 * Mimnimum of the total score
 *
 * ANALYSIS:
 * Think from the last step:
 * There are so many different ways to merge the stones
 * But at the end you will have only two piles
 * And you will just merge the two piles into one
 * Getting the score = all stones add up together
 *
 * Thus:
 *   - the score you get in the last-time merge is known
 *
 * Now think a bit more:
 * Each pile was originally merged from two smaller piles
 * With a score = all stones of left pile and right pile added up together
 *
 * Deduce the formula:
 *
 * possibilities of final scores of a list of N stones =
 *  sum of all stones in the list
 *    +
 *  possibilities of the sum of the final scores of merging each of its two sublists,
 *  where one list has X stones and theother list has Y stones
 *
 *  and you can go on and on...
 *
 * so we can see this is a recursive function
 * that keeps processing a list of N stones
 * and inside the function you need a loop
 * to iterate the {(x, y) | x + y = N}
 *
 * and our target is to find the smallest in this greate searching space.
 *
 * With this in mind we know how to test the correctness now.
 * However, this search is not optimal in terms of performance.
 *
 * We just want the minimal, let's optimize:
 *  the minimal score of merging a list of N stones =
 *    sum of all stones in the list
 *      +
 *    minimal score of possibilities of the sum of the final scores
 *    of merging each of its two sublists, where one list has X stones
 *    and the other list has Y stones
 *
 */

public class Solution {
    /**
     * @param A: An integer array
     * @return: An integer
     */
    public int stoneGame(int[] A) {

        if (A == null || A.length <= 0) return 0;

        this.stones = A;
        this.cache = new int[A.length][A.length];

        int start = 0;
        int end = A.length - 1;
        return getScore(start, end);
    }

    private int[][] cache = null;

    private int getScore(int start, int end) {
        // cache logic
        if (this.cache[start][end] != 0) return this.cache[start][end];

        // real business logic
        if (start >= end) return 0;
        int rangeSum = sumRange(start, end);
        boolean isAdjacent = (start + 1) == end;
        int answer = 0;
        if (isAdjacent) {
            answer = rangeSum;
        } else {
            int minMergeScore = minScore(start, end);
            answer = rangeSum + minMergeScore;
        }

        // cache and return answer
        this.cache[start][end] = answer;
        return answer;
    }

    private int minScore(int start, int end) {
        int min = Integer.MAX_VALUE;
        for(int x = start; x < end; x++) {
            int score = getScore(start, x) + getScore(x + 1, end);
            min = Math.min(min, score);
        }
        return min;
    }

    private int[] stones = null;
    private int sumRange(int start, int end) {
        int sum = 0;
        for(int i = start; i <= end; i++) {
            sum += this.stones[i];
        }
        return sum;
    }

}