Boats to Save People
<p>You are given an array <code>people</code> where <code>people[i]</code> is the weight of the <code>i<sup>th</sup></code> person, and an <strong>infinite number of boats</strong> where each boat can carry a maximum weight of <code>limit</code>. Each boat carries at most two people at the same time, provided the sum of the weight of those people is at most <code>limit</code>.</p>

<p>Return <em>the minimum number of boats to carry every given person</em>.</p>

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

<pre>
<strong>Input:</strong> people = [1,2], limit = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> 1 boat (1, 2)
</pre>

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

<pre>
<strong>Input:</strong> people = [3,2,2,1], limit = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> 3 boats (1, 2), (2) and (3)
</pre>

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

<pre>
<strong>Input:</strong> people = [3,5,3,4], limit = 5
<strong>Output:</strong> 4
<strong>Explanation:</strong> 4 boats (3), (3), (4), (5)
</pre>

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

<ul>
	<li><code>1 &lt;= people.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= people[i] &lt;= limit &lt;= 3 * 10<sup>4</sup></code></li>
</ul>
Greedy; Array; Two Pointers; Sorting
class Solution {
public:
    int numRescueBoats(vector<int>& people, int limit) {
        sort(people.begin(), people.end());
        int ans = 0;
        for (int i = 0, j = people.size() - 1; i <= j; --j) {
            if (people[i] + people[j] <= limit) {
                ++i;
            }
            ++ans;
        }
        return ans;
    }
};