Count Number of Teams
<p>There are <code>n</code> soldiers standing in a line. Each soldier is assigned a <strong>unique</strong> <code>rating</code> value.</p>

<p>You have to form a team of 3 soldiers amongst them under the following rules:</p>

<ul>
	<li>Choose 3 soldiers with index (<code>i</code>, <code>j</code>, <code>k</code>) with rating (<code>rating[i]</code>, <code>rating[j]</code>, <code>rating[k]</code>).</li>
	<li>A team is valid if: (<code>rating[i] &lt; rating[j] &lt; rating[k]</code>) or (<code>rating[i] &gt; rating[j] &gt; rating[k]</code>) where (<code>0 &lt;= i &lt; j &lt; k &lt; n</code>).</li>
</ul>

<p>Return the number of teams you can form given the conditions. (soldiers can be part of multiple teams).</p>

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

<pre>
<strong>Input:</strong> rating = [2,5,3,4,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can form three teams given the conditions. (2,3,4), (5,4,1), (5,3,1). 
</pre>

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

<pre>
<strong>Input:</strong> rating = [2,1,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong> We can&#39;t form any team given the conditions.
</pre>

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

<pre>
<strong>Input:</strong> rating = [1,2,3,4]
<strong>Output:</strong> 4
</pre>

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

<ul>
	<li><code>n == rating.length</code></li>
	<li><code>3 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= rating[i] &lt;= 10<sup>5</sup></code></li>
	<li>All the integers in <code>rating</code> are <strong>unique</strong>.</li>
</ul>
Binary Indexed Tree; Segment Tree; Array; Dynamic Programming
class Solution {
public:
    int numTeams(vector<int>& rating) {
        int n = rating.size();
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int l = 0, r = 0;
            for (int j = 0; j < i; ++j) {
                if (rating[j] < rating[i]) {
                    ++l;
                }
            }
            for (int j = i + 1; j < n; ++j) {
                if (rating[j] > rating[i]) {
                    ++r;
                }
            }
            ans += l * r;
            ans += (i - l) * (n - i - 1 - r);
        }
        return ans;
    }
};