Minimum Cost Tree From Leaf Values

Given an array arr of positive integers, consider all binary trees such that:

• Each node has either 0 or 2 children;
• The values of arr correspond to the values of each leaf in an in-order traversal of the tree.  (Recall that a node is a leaf if and only if it has 0 children.)
• The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree respectively.

Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node.  It is guaranteed this sum fits into a 32-bit integer.

 

Example 1:

Input: arr = [6,2,4]
Output: 32
Explanation:
There are two possible trees.  The first has non-leaf node sum 36, and the second has non-leaf node sum 32.

    24            24
   /  \          /  \
  12   4        6    8
 /  \               / \
6    2             2   4

思路：这题最标准的解法是dp;

dp[i,j] = Max(arr[i....k]) * Max(arr[k+1, j]) + dp[i, k] + dp[k+1, j]; 面试的时候，能相处这个已经很牛逼 了，更牛逼的是，这题可以换一种思维方式：

Let's review the problem again.

When we build a node in the tree, we compared the two numbers a and b.
In this process,
the smaller one is removed and we won't use it anymore,
and the bigger one actually stays.

The problem can translated as following:
Given an array A, choose two neighbors in the array a and b,
we can remove the smaller one min(a,b) and the cost is a * b.
What is the minimum cost to remove the whole array until only one left?

To remove a number a, it needs a cost a * b, where b >= a.
So a has to be removed by a bigger number.
We want minimize this cost, so we need to minimize b.

b has two candidates, the first bigger number on the left,
the first bigger number on the right.

The cost to remove a is a * min(left, right).

这题可以变成，每次弹走最小值，由最小值的左右两边比它大的最小值来计算cost，那么求左右两边最大值，就是单调递减栈；O(N）

class Solution {
    public int mctFromLeafValues(int[] arr) {
        if(arr == null || arr.length == 0) {
            return 0;
        }
        
        int n = arr.length;
        int[][] dp = new int[n][n];
        int[][] maxarr = new int[n][n];
        for(int i = 0; i < n; i++) {
            int localmax = 0;
            for(int j = i; j < n; j++) {
                if(arr[j] > localmax) {
                    localmax = arr[j];
                }
                maxarr[i][j] = localmax;
            }
        }
        
        for(int len = 1; len < n; len++) {
            for(int i = 0; i + len < n; i++) {
                int j = i + len;
                dp[i][j] = Integer.MAX_VALUE;
                if(len == 1) {
                    dp[i][j] = arr[i] * arr[j];
                } else {
                     for(int k = i; k < j; k++) {
                        dp[i][j] = Math.min(dp[i][j], 
                                          maxarr[i][k] * maxarr[k+1][j] + dp[i][k] + dp[k+1][j]);
                    }
                }
            }
        }
        
        return dp[0][n-1];
    }
}

class Solution {
    public int mctFromLeafValues(int[] arr) {
        if(arr == null || arr.length == 0) {
            return 0;
        }
        int res = 0;
        Stack<Integer> stack = new Stack<Integer>();
        stack.push(Integer.MAX_VALUE); // prevent empty stack.
        for(int a: arr) {
            while(stack.peek() <= a) {
                int mid = stack.pop();
                res += mid * Math.min(stack.peek(), a);
            }
            stack.push(a);
        }
        // stack contains Integer.MAX_VALUE;
        while(stack.size() > 2) {
            res += stack.pop() * stack.peek();
        }
        return res;
    }
}