Maximum Profitable Triplets With Increasing Prices II
<p>Given the <strong>0-indexed</strong> arrays <code>prices</code> and <code>profits</code> of length <code>n</code>. There are <code>n</code> items in an store where the <code>i<sup>th</sup></code> item has a price of <code>prices[i]</code> and a profit of <code>profits[i]</code>.</p>

<p>We have to pick three items with the following condition:</p>

<ul>
	<li><code>prices[i] &lt; prices[j] &lt; prices[k]</code> where <code>i &lt; j &lt; k</code>.</li>
</ul>

<p>If we pick items with indices <code>i</code>, <code>j</code> and <code>k</code> satisfying the above condition, the profit would be <code>profits[i] + profits[j] + profits[k]</code>.</p>

<p>Return<em> the <strong>maximum profit</strong> we can get, and </em><code>-1</code><em> if it&#39;s not possible to pick three items with the given condition.</em></p>

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

<pre>
<strong>Input:</strong> prices = [10,2,3,4], profits = [100,2,7,10]
<strong>Output:</strong> 19
<strong>Explanation:</strong> We can&#39;t pick the item with index i=0 since there are no indices j and k such that the condition holds.
So the only triplet we can pick, are the items with indices 1, 2 and 3 and it&#39;s a valid pick since prices[1] &lt; prices[2] &lt; prices[3].
The answer would be sum of their profits which is 2 + 7 + 10 = 19.</pre>

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

<pre>
<strong>Input:</strong> prices = [1,2,3,4,5], profits = [1,5,3,4,6]
<strong>Output:</strong> 15
<strong>Explanation:</strong> We can select any triplet of items since for each triplet of indices i, j and k such that i &lt; j &lt; k, the condition holds.
Therefore the maximum profit we can get would be the 3 most profitable items which are indices 1, 3 and 4.
The answer would be sum of their profits which is 5 + 4 + 6 = 15.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> prices = [4,3,2,1], profits = [33,20,19,87]
<strong>Output:</strong> -1
<strong>Explanation:</strong> We can&#39;t select any triplet of indices such that the condition holds, so we return -1.
</pre>

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

<ul>
	<li><code>3 &lt;= prices.length == profits.length &lt;= 50000</code></li>
	<li><code>1 &lt;= prices[i] &lt;= 5000</code></li>
	<li><code>1 &lt;= profits[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Binary Indexed Tree; Segment Tree; Array
class BinaryIndexedTree {
private:
    int n;
    vector<int> c;

public:
    BinaryIndexedTree(int n) {
        this->n = n;
        c.resize(n + 1, 0);
    }

    void update(int x, int v) {
        while (x <= n) {
            c[x] = max(c[x], v);
            x += x & -x;
        }
    }

    int query(int x) {
        int mx = 0;
        while (x > 0) {
            mx = max(mx, c[x]);
            x -= x & -x;
        }
        return mx;
    }
};

class Solution {
public:
    int maxProfit(vector<int>& prices, vector<int>& profits) {
        int n = prices.size();
        vector<int> left(n, 0);
        vector<int> right(n, 0);
        int m = *max_element(prices.begin(), prices.end());
        BinaryIndexedTree tree1(m + 1);
        BinaryIndexedTree tree2(m + 1);
        for (int i = 0; i < n; ++i) {
            int x = prices[i];
            left[i] = tree1.query(x - 1);
            tree1.update(x, profits[i]);
        }
        for (int i = n - 1; i >= 0; --i) {
            int x = m + 1 - prices[i];
            right[i] = tree2.query(x - 1);
            tree2.update(x, profits[i]);
        }
        int ans = -1;
        for (int i = 0; i < n; ++i) {
            if (left[i] > 0 && right[i] > 0) {
                ans = max(ans, left[i] + profits[i] + right[i]);
            }
        }
        return ans;
    }
};