Boats to Save People

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)


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


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.


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

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In a binary min heap, the minimum element can be found in:

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