<p>You may recall that an array <code>arr</code> is a <strong>mountain array</strong> if and only if:</p>

<ul>
	<li><code>arr.length &gt;= 3</code></li>
	<li>There exists some index <code>i</code> (<strong>0-indexed</strong>) with <code>0 &lt; i &lt; arr.length - 1</code> such that:
	<ul>
		<li><code>arr[0] &lt; arr[1] &lt; ... &lt; arr[i - 1] &lt; arr[i]</code></li>
		<li><code>arr[i] &gt; arr[i + 1] &gt; ... &gt; arr[arr.length - 1]</code></li>
	</ul>
	</li>
</ul>

<p>Given an integer array <code>nums</code>​​​, return <em>the <strong>minimum</strong> number of elements to remove to make </em><code>nums<em>​​​</em></code><em> </em><em>a <strong>mountain array</strong>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,1]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The array itself is a mountain array so we do not need to remove any elements.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,1,1,5,6,2,3,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One solution is to remove the elements at indices 0, 1, and 5, making the array nums = [1,5,6,3,1].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li>It is guaranteed that you can make a mountain array out of <code>nums</code>.</li>
</ul>

class Solution {
public:
    int dpToEnd[1001][1001];
    int dpToHead[1001][1001];
    int toEnd(const vector<int>& nList,int prevIdx,int idx){
        if(idx >= nList.size()){
            return 0;
        }
        if(dpToEnd[prevIdx][idx] != -1){
            return dpToEnd[prevIdx][idx];
        }
        int op1 = toEnd(nList,prevIdx,idx+1);
        
        int op2 = 0;
        if(nList[idx] < nList[prevIdx]){
            op2 = 1 + toEnd(nList,idx,idx+1);
        }
        
        int ret = max(op1,op2);
        dpToEnd[prevIdx][idx] = ret;
        return ret;
    }
    int toHead(const vector<int>& nList,int backIdx,int idx){
        if(idx<0){
            return 0;
        }
        if(dpToHead[backIdx][idx] != -1){
            return dpToHead[backIdx][idx];
        }
        int op1 = toHead(nList,backIdx,idx-1);
        int op2 = 0;
        if(nList[idx] < nList[backIdx]){
            op2 = 1 + toHead(nList,idx,idx-1);
        }
        int ret = max(op1,op2);
        dpToHead[backIdx][idx] = ret;
        return ret;
    }
    int minimumMountainRemovals(vector<int>& nums) {
        memset(dpToEnd,-1,sizeof(dpToEnd));
        memset(dpToHead,-1,sizeof(dpToHead));
        int ret = 0;
        for(int i = 1;i < nums.size()-1; ++i){
            int left = toHead(nums,i,i-1);
            int right = toEnd(nums,i,i+1);
            if(left==0 || right==0)
                continue;
            ret = max(ret,left+right+1);
        }
        return nums.size() - ret;
    }
};
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