 # LeetCode Boats to Save People Solution

You are given an array `people` where `people[i]` is the weight of the `ith` person, and an infinite number of boats where each boat can carry a maximum weight of `limit`. Each boat carries at most two people at the same time, provided the sum of the weight of those people is at most `limit`.

Return the minimum number of boats to carry every given person.

Example 1:

Input: `people = [1,2], limit = 3`
Output: `1`
Explanation: 1 boat (1, 2)

Example 2:

Input: `people = [3,2,2,1], limit = 3`
Output: `3`
Explanation: 3 boats (1, 2), (2) and (3)

Example 3:

Input: `people = [3,5,3,4], limit = 5`
Output: `4`
Explanation: 4 boats (3), (3), (4), (5)

Constraints:

• `1 <= people.length <= 5 * 104`
• `1 <= people[i] <= limit <= 3 * 104`

## Solution

We wish to perform two pointers in opposite directions on a sorted array `people`. To optimize the number of boats, we can assign a light weight person with a heavy person. Just as Two Sum Sorted try to pair two numbers adding up to `sum`, we try to pair two people with weight less than or equal to `limit`. If their total weight is more than the limit, then we can only assign the person with the heavier weight (`person[r]`) onto the boat so that the lighter person may be paired with someone else.

#### Implementation

``````1def numRescueBoats(self, people: List[int], limit: int) -> int:
2      people.sort()
3      l, r = 0 , len(people)-1
4      boat = 0
5      while l <= r:
6          if people[l] + people[r] <= limit:
7              l += 1
8          boat += 1
9          r -= 1
10      return boat``````