<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>arr</code>, return <em>the length of the longest subarray, which is a mountain</em>. Return <code>0</code> if there is no mountain subarray.</p>

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

<pre>
<strong>Input:</strong> arr = [2,1,4,7,3,2,5]
<strong>Output:</strong> 5
<strong>Explanation:</strong> The largest mountain is [1,4,7,3,2] which has length 5.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,2,2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no mountain.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong></p>

<ul>
	<li>Can you solve it using only one pass?</li>
	<li>Can you solve it in <code>O(1)</code> space?</li>
</ul>

class Solution {
public:
    int dp1[100100];
    int dp2[100100];

    int f1(int ind, vector<int>& arr) {
        // Base case:
        if (ind == 0) {
            return 0;
        }

        // Cache:
        if (dp1[ind] != -1) {
            return dp1[ind];
        }

        // Transition:
        int ans = 0;
        if (arr[ind] > arr[ind - 1]) {
            ans = 1 + f1(ind - 1, arr);
        }
        return dp1[ind] = ans;
    }

    int f2(int ind, vector<int>& arr) {
        // Base case:
        if (ind == arr.size() - 1) {
            return 0;
        }

        // Cache:
        if (dp2[ind] != -1) {
            return dp2[ind];
        }

        // Transition:
        int ans = 0;
        if (arr[ind] > arr[ind + 1]) {
            ans = 1 + f2(ind + 1, arr);
        }
        return dp2[ind] = ans;
    }

    int longestMountain(vector<int>& arr) {
        int n = arr.size();
        if (n < 3) return 0;

        // state : 
        // dp1[i] stores num of elem smaller than a[i] to its left
        // dp2[i] stores num of elem greater than a[i] to its right

        // final ans : 
        // max(dp1[i] + dp2[i] + 1)

        memset(dp1, -1, sizeof(dp1));
        memset(dp2, -1, sizeof(dp2));

        for (int i = 0; i < n; i++) {
            f1(i, arr);
        }

        for (int i = n - 1; i >= 0; i--) {
            f2(i, arr);
        }

        int maxi = 0;
        for (int i = 1; i < n - 1; i++) {
            if (dp1[i] > 0 && dp2[i] > 0) { 
                maxi = max(maxi, dp1[i] + dp2[i] + 1);
            }
        }

        return maxi < 3 ? 0 : maxi;
    }
};
22000