LintCode-Minimum Adjustment Cost

Given an integer array, adjust each integers so that the difference of every adjcent integers are not greater than a given number target.

If the array before adjustment is A, the array after adjustment is B, you should minimize the sum of |A[i]-B[i]| 

Note

You can assume each number in the array is a positive integer and not greater than 100

Example

Given [1,4,2,3] and target=1, one of the solutions is [2,3,2,3], the adjustment cost is 2 and it's minimal. Return 2.

Analysis:

Since there is boundary on the value of each element, we use dp to check every possible value of every element.

Solution:

public class Solution {
    /**
     * @param A: An integer array.
     * @param target: An integer.
     */
    public int MinAdjustmentCost(ArrayList<Integer> A, int target) {
        if (A.size() < 2) return 0;
    
        int[][] cost = new int[A.size()][100];
        for (int i = 0; i < 100; i++)
            cost[0][i] = Math.abs(A.get(0)-(i+1));
        
        for (int i=1;i<A.size();i++)
            for (int j=0;j<100;j++){
                cost[i][j]=Integer.MAX_VALUE;
                int diff = Math.abs(A.get(i)-(j+1));
                int upper = Math.min(j+target,99);
                int lower = Math.max(j-target,0);
                for (int k=lower;k<=upper;k++)
                    if (cost[i-1][k]+diff<cost[i][j])
                        cost[i][j]=cost[i-1][k]+diff;
            }
        int res = Integer.MAX_VALUE;
        for (int i=0;i<100;i++)
            if (cost[A.size()-1][i]<res)
                res = cost[A.size()-1][i];
        return res;
    }
}

NOTE 1: we only need to use two single dimension arrays, because cost[i][j] is only related with cost[i-1][j].

NOTE 2: if we need to output the new array, we need a list of arrays to record the current array of cost[i][j], it equals to {list of cost[i][k]} + {j}.